These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance
In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
That’s the beauty of math: There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are seven more brutally difficult math problems that once seemed impossible until mathematicians found a breakthrough.
These ten brutally difficult math problems once seemed impossible until mathematicians eventually solved them—even if it took them years, decades, or centuries.
These Are the 10 Hardest Math Problems Ever Solved
They’re guaranteed to make your head spin.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
That’s the beauty of math : There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are nine more brutally difficult math problems that once seemed impossible, until mathematicians found a breakthrough.
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The Poincaré Conjecture
In 2000, the Clay Mathematics Institute , a non-profit dedicated to “increasing and disseminating mathematical knowledge,” asked the world to solve seven math problems and offered $1,000,000 to anybody who could crack even one. Today, they’re all still unsolved, except for the Poincaré conjecture.
Henri Poincaré was a French mathematician who, around the turn of the 20th century, did foundational work in what we now call topology. Here’s the idea: Topologists want mathematical tools for distinguishing abstract shapes. For shapes in 3D space, like a ball or a donut, it wasn’t very hard to classify them all . In some significant sense, a ball is the simplest of these shapes.
Poincaré then went up to 4-dimensional stuff, and asked an equivalent question. After some revisions and developments, the conjecture took the form of “Every simply-connected, closed 3-manifold is homeomorphic to S^3,” which essentially says “the simplest 4D shape is the 4D equivalent of a sphere.”
Still with us?
A century later, in 2003, a Russian mathematician named Grigori Perelman posted a proof of Poincaré’s conjecture on the modern open math forum arXiv. Perelman’s proof had some small gaps, and drew directly from research by American mathematician Richard Hamilton. It was groundbreaking, yet modest.
After the math world spent a few years verifying the details of Perelman’s work, the awards began . Perelman was offered the million-dollar Millennium Prize, as well as the Fields Medal, often called the Nobel Prize of Math. Perelman rejected both. He said his work was for the benefit of mathematics, not personal gain, and also that Hamilton, who laid the foundations for his proof, was at least as deserving of the prizes.
Fermat’s Last Theorem
Pierre de Fermat was a 17th-century French lawyer and mathematician. Math was apparently more of a hobby for Fermat, and so one of history’s greatest math minds communicated many of his theorems through casual correspondence. He made claims without proving them, leaving them to be proven by other mathematicians decades, or even centuries, later. The most challenging of these has become known as Fermat’s Last Theorem.
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
Fermat famously wrote the Last Theorem by hand in the margin of a textbook, along with the comment that he had a proof, but could not fit it in the margin. For centuries, the math world has been left wondering if Fermat really had a valid proof in mind.
Flash forward 330 years after Fermat’s death to 1995, when British mathematician Sir Andrew Wiles finally cracked one of history’s oldest open problems . For his efforts, Wiles was knighted by Queen Elizabeth II and was awarded a unique honorary plaque in lieu of the Fields Medal, since he was just above the official age cutoff to receive a Fields Medal.
Wiles managed to combine new research in very different branches of math in order to solve Fermat’s classic number theory question. One of these topics, Elliptic Curves, was completely undiscovered in Fermat’s time, leading many to believe Fermat never really had a proof of his Last Theorem.
The Classification of Finite Simple Groups
From solving Rubik’s Cube to proving a fact about body-swapping on Futurama , abstract algebra has a wide range of applications. Algebraic groups are sets that follow a few basic properties, like having an “identity element,” which works like adding 0.
Groups can be finite or infinite, and if you want to know what groups of a particular size n look like, it can get very complicated depending on your choice of n .
If n is 2 or 3, there’s only one way that group can look. When n hits 4, there are two possibilities. Naturally, mathematicians wanted a comprehensive list of all possible groups for any given size.
The complete list took decades to finish conclusively, because of the difficulties in being sure that it was indeed complete. It’s one thing to describe what infinitely many groups look like, but it’s even harder to be sure the list covers everything. Arguably the greatest mathematical project of the 20th century, the classification of finite simple groups was orchestrated by Harvard mathematician Daniel Gorenstein, who in 1972 laid out the immensely complicated plan.
By 1985, the work was nearly done, but spanned so many pages and publications that it was unthinkable for one person to peer review. Part by part, the many facets of the proof were eventually checked and the completeness of the classification was confirmed.
By the 1990s, the proof was widely accepted. Subsequent efforts were made to streamline the titanic proof to more manageable levels, and that project is still ongoing today .
The Four Color Theorem
This one is as easy to state as it is hard to prove.
Grab any map and four crayons. It’s possible to color each state (or country) on the map, following one rule: No states that share a border get the same color.
The fact that any map can be colored with five colors—the Five Color Theorem —was proven in the 19th century. But getting that down to four took until 1976.
Two mathematicians at the University of Illinois, Urbana-Champaign, Kenneth Appel and Wolfgang Hakan, found a way to reduce the proof to a large, finite number of cases . With computer assistance, they exhaustively checked the nearly 2,000 cases, and ended up with an unprecedented style of proof.
Arguably controversial since it was partially conceived in the mind of a machine, Appel and Hakan’s proof was eventually accepted by most mathematicians. It has since become far more common for proofs to have computer-verified parts, but Appel and Hakan blazed the trail.
(The Independence of) The Continuum Hypothesis
In the late 19th century, a German mathematician named Georg Cantor blew everyone’s minds by figuring out that infinities come in different sizes, called cardinalities. He proved the foundational theorems about cardinality, which modern day math majors tend to learn in their Discrete Math classes.
Cantor proved that the set of real numbers is larger than the set of natural numbers, which we write as |ℝ|>|ℕ|. It was easy to establish that the size of the natural numbers, |ℕ|, is the first infinite size; no infinite set is smaller than ℕ.
Now, the real numbers are larger, but are they the second infinite size? This turned out to be a much harder question, known as The Continuum Hypothesis (CH) .
If CH is true, then |ℝ| is the second infinite size, and no infinite sets are smaller than ℝ, yet larger than ℕ. And if CH is false, then there is at least one size in between.
So what’s the answer? This is where things take a turn.
CH has been proven independent, relative to the baseline axioms of math. It can be true, and no logical contradictions follow, but it can also be false, and no logical contradictions will follow.
It’s a weird state of affairs, but not completely uncommon in modern math. You may have heard of the Axiom of Choice, another independent statement. The proof of this outcome spanned decades and, naturally, split into two major parts: the proof that CH is consistent, and the proof that the negation of CH is consistent.
The first half is thanks to Kurt Gödel, the legendary Austro-Hungarian logician. His 1938 mathematical construction, known as Gödel’s Constructible Universe , proved CH compatible with the baseline axioms, and is still a cornerstone of Set Theory classes. The second half was pursued for two more decades until Paul Cohen, a mathematician at Stanford, solved it by inventing an entire method of proof in Model Theory known as “forcing.”
Gödel’s and Cohen’s halves of the proof each take a graduate level of Set Theory to approach, so it’s no wonder this unique story has been esoteric outside mathematical circles.
Gödel’s Incompleteness Theorems
Gödel’s work in mathematical logic was totally next-level. On top of proving stuff, Gödel also liked to prove whether or not it was possible to prove stuff . His Incompleteness Theorems are often misunderstood, so here’s a perfect chance to clarify them.
Gödel’s First Incompleteness Theorem says that, in any proof language, there are always unprovable statements. There’s always something that’s true, that you can’t prove true. It’s possible to understand a (non-mathematically rigorous) version of Gödel’s argument, with some careful thinking. So buckle up, here it is: Consider the statement, “This statement cannot be proven true.”
Think through every case to see why this is an example of a true, but unprovable statement. If it’s false, then what it says is false, so then it can be proven true, which is contradictory, so this case is impossible. On the other extreme, if it did have a proof, then that proof would prove it true … making it true that it has no proof, which is contradictory, killing this case. So we’re logically left with the case that the statement is true, but has no proof. Yeah, our heads are spinning, too.
But follow that nearly-but-not-quite-paradoxical trick, and you’ve illustrated that Gödel’s First Incompleteness Theorem holds.
Gödel’s Second Incompleteness Theorem is similarly weird. It says that mathematical “formal systems” can’t prove themselves consistent. A consistent system is one that won’t give you any logical contradictions.
Here’s how you can think of that. Imagine Amanda and Bob each have a set of mathematical axioms—baseline math rules—in mind. If Amanda can use her axioms to prove that Bob’s axiom system is free of contradictions, then it’s impossible for Bob to use his axioms to prove Amanda’s system doesn’t yield contradictions.
So when mathematicians debate the best choices for the essential axioms of mathematics (it’s much more common than you might imagine) it’s crucial to be aware of this phenomenon.
The Prime Number Theorem
There are plenty of theorems about prime numbers . One of the simplest facts—that there are infinitely many prime numbers—can even be adorably fit into haiku form .
The Prime Number Theorem is more subtle; it describes the distribution of prime numbers along the number line. More precisely, it says that, given a natural number N, the number of primes below N is approximately N/log(N) ... with the usual statistical subtleties to the word “approximately” there.
Drawing on mid-19th-century ideas, two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, independently proved the Prime Number Theorem in 1898. Since then, the proof has been a popular target for rewrites, enjoying many cosmetic revisions and simplifications. But the impact of the theorem has only grown.
The usefulness of the Prime Number Theorem is huge. Modern computer programs that deal with prime numbers rely on it. It’s fundamental to primality testing methods, and all the cryptology that goes with that.
Solving Polynomials by Radicals
Remember the quadratic formula ? Given ax²+bx+c=0, the solution is x=(-b±√(b^2-4ac))/(2a), which may have felt arduous to memorize in high school, but you have to admit is a conveniently closed-form solution.
Now, if we go up to ax³+bx²+cx+d=0, a closed form for “x=” is possible to find, although it’s much bulkier than the quadratic version. It’s also possible, yet ugly, to do this for degree 4 polynomials ax⁴+bx³+cx²+dx+f=0.
The goal of doing this for polynomials of any degree was noted as early as the 15th century. But from degree 5 on, a closed form is not possible. Writing the forms when they’re possible is one thing, but how did mathematicians prove it’s not possible from 5 up?
The world was only starting to comprehend the brilliance of French mathematician Evariste Galois when he died at the age of 20 in 1832. His life included months spent in prison, where he was punished for his political activism, writing ingenious, yet unrefined mathematics to scholars, and it ended in a fatal duel.
Galois’ ideas took decades after his death to be fully understood, but eventually they developed into an entire theory now called Galois Theory . A major theorem in this theory gives exact conditions for when a polynomial can be “solved by radicals,” meaning it has a closed form like the quadratic formula. All polynomials up to degree 4 satisfy these conditions, but starting at degree 5, some don’t, and so there’s no general form for a solution for any degree higher than 4.
Trisecting an Angle
To finish, let’s go way back in history.
The Ancient Greeks wondered about constructing lines and shapes in various ratios, using the tools of an unmarked compass and straightedge . If someone draws an angle on some paper in front of you, and gives you an unmarked ruler, a basic compass, and a pen, it’s possible for you to draw the line that cuts that angle exactly in half. It’s a quick four steps, nicely illustrated like this , and the Greeks knew it two millennia ago.
What eluded them was cutting an angle in thirds. It stayed elusive for literally 15 centuries, with hundreds of attempts in vain to find a construction. It turns out such a construction is impossible.
Modern math students learn the angle trisection problem—and how to prove it’s not possible—in their Galois Theory classes. But, given the aforementioned period of time it took the math world to process Galois’ work, the first proof of the problem was due to another French mathematician, Pierre Wantzel . He published his work in 1837, 16 years after the death of Galois, but nine years before most of Galois’ work was published.
Either way, their insights are similar, casting the construction question into one about properties of certain representative polynomials. Many other ancient construction questions became approachable with these methods, closing off some of the oldest open math questions in history.
So if you ever time-travel to ancient Greece, you can tell them their attempts at the angle trisection problem are futile.
Dave Linkletter is a Ph.D. candidate in Pure Mathematics at the University of Nevada, Las Vegas. His research is in Large Cardinal Set Theory. He also teaches undergrad classes, and enjoys breaking down popular math topics for wide audiences.
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The fascination and complexity of the world's hardest math problems
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Mathematics has been a fascinating and challenging subject for centuries. From the ancient Greeks to modern-day mathematicians, the pursuit of understanding and mastering math has been a source of intrigue and intellectual curiosity.
But have you ever wondered what the hardest math problem is? What could be so challenging and complex that even the most brilliant mathematicians have yet to find a solution?
This article will explore some of the hardest math problems ever posed and the different approaches mathematicians have used to solve these problems.
So, buckle up and get ready to explore some of the most challenging math problems ever!
5 hardest math problems in the world
Mathematics has been around for thousands of years and has contributed to numerous fields, including science, engineering, and finance. However, some math problems have stumped even the most brilliant mathematicians for centuries.
Here are some brutally difficult math problems that once seemed impossible to solve and some that still are.
The Poincaré Conjecture
The Poincaré Conjecture, proposed by mathematician Henri Poincaré in 1904, is a problem that stumped the mathematics community for nearly 100 years.
It states that every connected, closed three-dimensional space is topologically equivalent to a three-dimensional sphere (S3).
To understand what this means, we need to delve into the world of topology. Topology is the study of the properties of objects that remain unchanged when they are stretched, bent, or otherwise deformed. In other words, topologists are interested in the ways that objects can be transformed without tearing or breaking.
The Poincaré Conjecture concerns the topology of three-dimensional spaces. A three-dimensional space is a space volume with three dimensions – length, width, and height. A sphere is a three-dimensional object with a round, curved surface.
The Poincaré Conjecture proposes that every simply-connected three-dimensional space (meaning it has no holes or voids) which is closed (meaning it has no edges or boundaries) is topologically equivalent to a three-sphere (S3) — the set of points in four-dimensional space at some fixed distance to a given point. This may seem simple, but it took over 100 years to fully prove the conjecture.
Poincaré later extended his conjecture to any dimension (n-sphere). In 1961, the American mathematician Stephen Smale showed that the conjecture is true for n ≥ 5; in 1983, the American mathematician Michael Freedman showed that it is true for n = 4, and in 2002, the Russian mathematician Grigori Perelman finally completed the solution by proving it true for n = 3.
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Perelman finally solved the problem using a combination of topology and geometry. All three mathematicians were awarded a Fields Medal , one of the highest honors in mathematics. Perelman refused his Fields Medal. He was also awarded a million-dollar prize by the Clay Mathematics Institute (CMI) of Cambridge, Mass., for solving one of the world's most difficult mathematical problems (seven problems dubbed the Millenium Problems), which he also refused.
The Poincaré Conjecture has had significant implications in the field of topology and has been described as the "holy grail" of mathematics. It has opened up new research avenues and inspired numerous mathematicians to tackle other challenging problems in the field.
The Riemann Hypothesis
The Riemann Hypothesis is a mathematical conjecture proposed by the German mathematician Bernhard Riemann in 1859 that has puzzled mathematicians for over 150 years.
It states that every nontrivial zero of the Riemann zeta function has a real part of ½.
The Riemann zeta function is one that can be used to represent the distribution of prime numbers. Prime numbers are only divisible by themselves and 1, such as 2, 3, 5, 7, and 11. The distribution of prime numbers has long been of interest to mathematicians, as understanding their patterns and relationships can lead to new insights into number theory and other areas of mathematics.
The Riemann Hypothesis suggests a relationship exists between the distribution of prime numbers and the zeros of the Riemann zeta function. If this relationship is proven to be accurate, it could have significant implications in the field of number theory and potentially lead to discoveries in other areas of mathematics.
Despite being considered one of the most important unsolved problems in mathematics, the Riemann Hypothesis is yet to be proven or disproven. Many mathematicians have attempted to solve it, but the conjecture remains elusive.
In 2002, mathematician Michael Atiyah claimed to have proof of the Riemann Hypothesis, but it is yet to be formally accepted by the mathematical community.
The hypothesis is another of the seven Millennium Prize Problems set by the Clay Institute. And anyone who can establish the validity or invalidity of the Riemann hypothesis will receive a prize of $1 million.
The Collatz Conjecture
The Collatz conjecture, also known as the "3n + 1" problem, is a mathematical problem that involves taking any positive integer and repeatedly applying a specific set of rules until you reach the number 1.
The rules are as follows:
1. If the number is even, divide it by 2.
2. If the number is odd, triple it and add 1.
For example, let's start with the number 7:
7 is odd, so we triple it and add 1 to get 22
22 is even, so we divide it by 2 to get 11
11 is odd, so we triple it and add 1 to get 34
34 is even, so we divide it by 2 to get 17
17 is odd, so we triple it and add 1 to get 52
52 is even, so we divide it by 2 to get 26
26 is even, so we divide it by 2 to get 13
13 is odd, so we triple it and add 1 to get 40
40 is even, so we divide it by 2 to get 20
20 is even, so we divide it by 2 to get 10
10 is even, so we divide it by 2 to get 5
5 is odd, so we triple it and add 1 to get 16
16 is even, so we divide it by 2 to get 8
8 is even, so we divide it by 2 to get 4
4 is even, so we divide it by 2 to get 2
2 is even, so we divide it by 2 to get 1
We have now reached the number 1, which means we can stop. This sequence of numbers we generated (7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1) is the Collatz sequence for the number 7.
The Collatz conjecture states that no matter which positive integer you start with, you will always eventually reach the number 1 if you follow these rules. In other words, the conjecture claims that the Collatz sequence for any positive integer will ultimately reach the number 1.
Despite many efforts, the Collatz conjecture has not yet been proven or disproven. It is considered one of the most famous unsolved problems in mathematics and has fascinated mathematicians for many years.
One exciting aspect of the Collatz conjecture is that it is very simple to understand and apply. Still, so far, people are yet to be able to solve it, even the most famous mathematicians.
In 2019, mathematician Terence Tao made a breakthrough in the problem, but he subsequently explained that this was only a partial solution.
The Collatz conjecture has also been studied in computer science, as it can be used to create efficient algorithms for specific types of calculations.
Fermat's Last Theorem
Fermat's Last Theorem, named after the French mathematician Pierre de Fermat , is a famous statement in mathematics, stating that there are no positive integers a, b, and c that satisfy the equation an + bn = cn for any integer value of n greater than 2.
In other words, it is impossible to find three integers that can be plugged into the equation an + bn = cn such that the equation is true for any value of n greater than 2.
Fermat first stated this theorem in the margin of a math book in 1637, but he never provided proof. The theorem remained unproven for over 350 years until Andrew Wiles, a mathematician at the University of Oxford, finally published a proof in 1994.
Fermat's Last Theorem has fascinated mathematicians for centuries because it is such a simple statement that seems to defy logic. It's hard to believe that there could be no solution to the equation an + bn = cn for any value of n greater than 2, but that's exactly what the theorem states.
So why was it so difficult to prove Fermat's Last Theorem? Part of the reason is that it involves a type of math called number theory, which deals with the properties of integers. Proving the theorem required a deep understanding of number theory and advanced mathematical techniques like elliptic curves and modular forms.
Despite the difficulty of proving Fermat's Last Theorem, it has significantly impacted mathematics. It has inspired many mathematicians to pursue careers in number theory and led to the development of new mathematical concepts and techniques.
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The Continuum Hypothesis
The Continuum Hypothesis is a mathematical problem involving the concept of infinity and the size of infinite sets. It was first proposed by Georg Cantor in 1878 and has remained one of the unsolvable and hardest math problems ever since.
The Continuum Hypothesis asks whether there is a set of numbers larger than natural numbers (1, 2, 3, etc.) but smaller than real numbers (e.g., all numbers on the number line). This set of numbers, if it exists, would be known as the "continuum."
One way to understand the Continuum Hypothesis is to consider the concept of "cardinality," which refers to the number of elements in a set. For example, the set of natural numbers has an infinite cardinality because it contains an infinite number of elements. The set of real numbers also has an infinite cardinality, but it is a larger infinity than the set of natural numbers.
The Continuum Hypothesis suggests that no set of numbers has an infinite cardinality between the set of natural numbers and the set of real numbers. In other words, it indicates that no set of numbers is "larger" than the set of natural numbers but "smaller" than the set of real numbers.
The Continuum Hypothesis has been the subject of much debate and controversy among mathematicians. Some have argued that it is simply a matter of definition – that the concept of an infinite set is too vague and ambiguous to be proven or disproven.
Others have attempted to prove or disprove the Continuum Hypothesis using various mathematical techniques, but no one has conclusively proven or disproved it.
In conclusion, the world's hardest math problems are indeed the cream of the crop when it comes to challenging the limits of human understanding and problem-solving skills. From the elusive Continuum Hypothesis to the mind-bending Riemann Hypothesis, these problems continue to stump even the most brilliant mathematicians.
But despite their difficulty, these problems continue to inspire and motivate mathematicians to push the boundaries of what is possible.
Whether or not they are ever solved, they serve as a testament to the boundless potential of the human mind and the ever-evolving nature of our understanding of the world around us.
While some of these problems may never be fully solved, they continue to inspire and drive progress in the field of mathematics and serve as a testament to the vast and mysterious nature of this subject.
As we saw in our recent article , even seemingly complex math problems can have practical and real-world implications.
So the next time you encounter a particularly challenging math problem, don't be discouraged – you may be on the path to solving one of the hardest math problems in the world!
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Cyborgs may have only been shown in Hollywood films up until this point, but a group of scientists from the RIKEN Cluster for Pioneering Research (CPR) in Japan has made them a reality.
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These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance
In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
That’s the beauty of math: There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are seven more brutally difficult math problems that once seemed impossible until mathematicians found a breakthrough.
1) The Poincaré Conjecture
In 2000, the Clay Mathematics Institute, a non-profit dedicated to “increasing and disseminating mathematical knowledge,” asked the world to solve seven math problems and offered $1,000,000 to anybody who could crack even one. Today, they’re all still unsolved, except for the Poincaré conjecture.
Henri Poincaré was a French mathematician who, around the turn of the 20th century, did foundational work in what we now call topology. Here’s the idea: Topologists want mathematical tools for distinguishing abstract shapes. For shapes in 3D space, like a ball or a donut, it wasn’t very hard to classify them all . In some significant sense, a ball is the simplest of these shapes.
Poincaré then went up to 4-dimensional stuff, and asked an equivalent question. After some revisions and developments, the conjecture took the form of “Every simply-connected, closed 3-manifold is homeomorphic to S^3,” which essentially says “the simplest 4D shape is the 4D equivalent of a sphere.”
Still with us?
A century later, in 2003, a Russian mathematician named Grigori Perelman posted a proof of Poincaré’s conjecture on the modern open math forum arXiv. Perelman’s proof had some small gaps, and drew directly from research by American mathematician Richard Hamilton. It was groundbreaking, yet modest.
After the math world spent a few years verifying the details of Perelman’s work, the awards began . Perelman was offered the million-dollar Millennium Prize, as well as the Fields Medal, often called the Nobel Prize of Math. Perelman rejected both. He said his work was for the benefit of mathematics, not personal gain, and also that Hamilton, who laid the foundations for his proof, was at least as deserving of the prizes.
2) Fermat's Last Theorem
Pierre de Fermat was a 17th-century French lawyer and mathematician. Math was apparently more of a hobby for Fermat, and so one of history’s greatest math minds communicated many of his theorems through casual correspondence. He made claims without proving them, leaving them to be proven by other mathematicians decades, or even centuries, later. The most challenging of these has become known as Fermat’s Last Theorem.
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
Fermat famously wrote the Last Theorem by hand in the margin of a textbook, along with the comment that he had a proof, but could not fit it in the margin. For centuries, the math world has been left wondering if Fermat really had a valid proof in mind.
Flash forward 330 years after Fermat’s death to 1995, when British mathematician Sir Andrew Wiles finally cracked one of history’s oldest open problems . For his efforts, Wiles was knighted by Queen Elizabeth II and was awarded a unique honorary plaque in lieu of the Fields Medal, since he was just above the official age cutoff to receive a Fields Medal.
Wiles managed to combine new research in very different branches of math in order to solve Fermat’s classic number theory question. One of these topics, Elliptic Curves, was completely undiscovered in Fermat’s time, leading many to believe Fermat never really had a proof of his Last Theorem.
3) The Four Color Theorem
This one is as easy to state as it is hard to prove.
Grab any map and four crayons. It’s possible to color each state (or country) on the map, following one rule: No states that share a border get the same color.
The fact that any map can be colored with five colors—the Five Color Theorem —was proven in the 19th century. But getting that down to four took until 1976.
Two mathematicians at the University of Illinois, Urbana-Champaign, Kenneth Appel and Wolfgang Hakan, found a way to reduce the proof to a large, finite number of cases . With computer assistance, they exhaustively checked the nearly 2,000 cases, and ended up with an unprecedented style of proof.
Arguably controversial since it was partially conceived in the mind of a machine, Appel and Hakan’s proof was eventually accepted by most mathematicians. It has since become far more common for proofs to have computer-verified parts, but Appel and Hakan blazed the trail.
4) (The Independence of) The Continuum Hypothesis
In the late 19th century, a German mathematician named Georg Cantor blew everyone’s minds by figuring out that infinities come in different sizes, called cardinalities. He proved the foundational theorems about cardinality, which modern day math majors tend to learn in their Discrete Math classes.
Cantor proved that the set of real numbers is larger than the set of natural numbers, which we write as |ℝ|>|ℕ|. It was easy to establish that the size of the natural numbers, |ℕ|, is the first infinite size; no infinite set is smaller than ℕ.
Now, the real numbers are larger, but are they the second infinite size? This turned out to be a much harder question, known as The Continuum Hypothesis (CH) .
If CH is true, then |ℝ| is the second infinite size, and no infinite sets are smaller than ℝ, yet larger than ℕ. And if CH is false, then there is at least one size in between.
So what’s the answer? This is where things take a turn.
CH has been proven independent, relative to the baseline axioms of math. It can be true, and no logical contradictions follow, but it can also be false, and no logical contradictions will follow.
It’s a weird state of affairs, but not completely uncommon in modern math. You may have heard of the Axiom of Choice, another independent statement. The proof of this outcome spanned decades and, naturally, split into two major parts: the proof that CH is consistent, and the proof that the negation of CH is consistent.
The first half is thanks to Kurt Gödel, the legendary Austro-Hungarian logician. His 1938 mathematical construction, known as Gödel’s Constructible Universe , proved CH compatible with the baseline axioms, and is still a cornerstone of Set Theory classes. The second half was pursued for two more decades until Paul Cohen, a mathematician at Stanford, solved it by inventing an entire method of proof in Model Theory known as “forcing.”
Gödel’s and Cohen’s halves of the proof each take a graduate level of Set Theory to approach, so it’s no wonder this unique story has been esoteric outside mathematical circles.
5) The Prime Number Theorem
There are plenty of theorems about prime numbers . One of the simplest facts—that there are infinitely many prime numbers—can even be adorably fit into haiku form .
The Prime Number Theorem is more subtle; it describes the distribution of prime numbers along the number line. More precisely, it says that, given a natural number N, the number of primes below N is approximately N/log(N) ... with the usual statistical subtleties to the word “approximately” there.
Drawing on mid-19th-century ideas, two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, independently proved the Prime Number Theorem in 1898. Since then, the proof has been a popular target for rewrites, enjoying many cosmetic revisions and simplifications. But the impact of the theorem has only grown.
The usefulness of the Prime Number Theorem is huge. Modern computer programs that deal with prime numbers rely on it. It’s fundamental to primality testing methods, and all the cryptology that goes with that.
6) Solving Polynomials by Radicals
Remember the quadratic formula ? Given ax²+bx+c=0, the solution is x=(-b±√(b^2-4ac))/(2a), which may have felt arduous to memorize in high school, but you have to admit is a conveniently closed-form solution.
Now, if we go up to ax³+bx²+cx+d=0, a closed form for “x=” is possible to find, although it’s much bulkier than the quadratic version. It’s also possible, yet ugly, to do this for degree 4 polynomials ax⁴+bx³+cx²+dx+f=0.
The goal of doing this for polynomials of any degree was noted as early as the 15th century. But from degree 5 on, a closed form is not possible. Writing the forms when they’re possible is one thing, but how did mathematicians prove it’s not possible from 5 up?
The world was only starting to comprehend the brilliance of French mathematician Evariste Galois when he died at the age of 20 in 1832. His life included months spent in prison, where he was punished for his political activism, writing ingenious, yet unrefined mathematics to scholars, and it ended in a fatal duel.
Galois’ ideas took decades after his death to be fully understood, but eventually they developed into an entire theory now called Galois Theory . A major theorem in this theory gives exact conditions for when a polynomial can be “solved by radicals,” meaning it has a closed form like the quadratic formula. All polynomials up to degree 4 satisfy these conditions, but starting at degree 5, some don’t, and so there’s no general form for a solution for any degree higher than 4.
7) Trisecting an Angle
To finish, let’s go way back in history.
The Ancient Greeks wondered about constructing lines and shapes in various ratios, using the tools of an unmarked compass and straightedge. If someone draws an angle on some paper in front of you, and gives you an unmarked ruler, a basic compass, and a pen, it’s possible for you to draw the line that cuts that angle exactly in half. It’s a quick four steps, nicely illustrated like this , and the Greeks knew it two millennia ago.
What eluded them was cutting an angle in thirds. It stayed elusive for literally 15 centuries, with hundreds of attempts in vain to find a construction. It turns out such a construction is impossible.
Modern math students learn the angle trisection problem—and how to prove it’s not possible—in their Galois Theory classes. But, given the aforementioned period of time it took the math world to process Galois’ work, the first proof of the problem was due to another French mathematician, Pierre Wantzel . He published his work in 1837, 16 years after the death of Galois, but nine years before most of Galois’ work was published.
Either way, their insights are similar, casting the construction question into one about properties of certain representative polynomials. Many other ancient construction questions became approachable with these methods, closing off some of the oldest open math questions in history.
So if you ever time-travel to ancient Greece, you can tell them their attempts at the angle trisection problem are futile.
These ten brutally difficult math problems once seemed impossible until mathematicians eventually solved them—even if it took them years, decades, or centuries.
Longest-standing maths problem (current)
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5 of the world’s toughest unsolved maths problems
The Open Problems in Mathematical Physics is a list of the most monstrous maths riddles in physics. Here are five of the top problems that remain unsolved
By Benjamin Skuse
7 February 2019
Mike Dunning/Getty
1. Separatrix Separation
A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?
Science History Images / Alamy Stock Photo
2. Navier–Stokes
The Navier-Stokes equations, developed in 1822, are used to describe the motion of viscous fluid. Things like air passing over an aircraft wing or water flowing out of a tap. But there are certain situations in which it is unclear whether the equations fail or give no answer at all. Many mathematicians have tried – and failed – to resolve the matter, including Mukhtarbay Otelbaev of the Eurasian National University in Astana, Kazakhstan. In 2014, he claimed a solution, but later retracted it. This is one problem that is worth more than just prestige. It is also one of the Millennium Prize Problems , which means anyone who solves it can claim $1 million in prize money.
Read more: The baffling quantum maths solution it took 10 years to understand
Cecile Lavabre/Getty
3. Exponents and dimensions
Imagine a squirt of perfume diffusing across a room. The movement of each molecule is random, a process called Brownian motion, even if the way the gas wafts overall is predictable. There is a mathematical language that can describe things like this, but not perfectly. It can provide exact solutions by bending its own rules or it can remain strict, but never quite arrive at the exact solution. Could it ever tick both boxes? That is what the exponents and dimensions problem asks. Apart from the quantum Hall conductance problem , this is the only one on the list that is at least partially solved. In 2000, Gregory Lawler, Oded Schramm and Wendelin Werner proved that exact solutions to two problems in Brownian motion can be found without bending the rules. It earned them a Fields medal, the maths equivalent of a Nobel prize. More recently, Stanislav Smirnov at the University of Geneva in Switzerland solved a related problem, which resulted in him being awarded the Fields medal in 2010.
Godong / Alamy Stock Photo
4. Impossibility theorems
There are plenty of mathematical expressions that have no exact solution. Take one of the most famous numbers ever, pi, which is the ratio of a circle’s circumference to its diameter. Proving that it was impossible for pi’s digits after the decimal point to ever end was one of the greatest contributions to maths. Physicists similarly say that it is impossible to find solutions to certain problems, like finding the exact energies of electrons orbiting a helium atom. But can we prove that impossibility?
Tetra Images/Getty
5. Spin glass
To understand this problem, you need to know about spin, a quantum mechanical property of atoms and particles like electrons, which underlies magnetism. You can think of it like an arrow that can point up or down. Electrons inside blocks of materials are happiest if they sit next to electrons that have the opposite spin, but there are some arrangements where that isn’t possible. In these frustrated magnets, spins often flip around randomly in a way that, it turns out, is a useful model of other disordered systems including financial markets. But we have limited ways of mathematically describing how systems like this behave. This spin glass question asks if we can find a good way of doing it.
• See the full list of unsolved problems: Open Problems in Mathematical Physics
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10 Math Equations That Have Never Been Solved
By Kathleen Cantor, 10 Sep 2020
Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?
Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.
1. The Riemann Hypothesis
Equation: σ (n) ≤ Hn +ln (Hn)eHn
- Where n is a positive integer
- Hn is the n-th harmonic number
- σ(n) is the sum of the positive integers divisible by n
For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?
This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the Clay Mathematics Foundation for its solution.
2. The Collatz Conjecture
Equation: 3n+1
- where n is a positive integer n/2
- where n is a non-negative integer
Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.
This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.
3. The Erdős-Strauss Conjecture
Equation: 4/n=1/a+1/b+1/c
- a, b and c are positive integers.
This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.
This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.
4. Equation Four
Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?
Looks pretty straight forward, does it? Here is a little context on the problem.
Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.
5. Goldbach's Conjecture
Equation: Prove that x + y = n
- where x and y are any two primes
This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.
If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.
6. Equation Six
Equation: Prove that (K)n = JK1N(q)JO1N(q)
- Where O = unknot (we are dealing with knot theory )
- (K)n = Kashaev's invariant of K for any K or knot
- JK1N(q) of K is equal to N- colored Jones polynomial
- We also have the volume of conjecture as (EQ3)
- Here vol(K) = hyperbolic volume
This equation tries to portray the relationship between quantum invariants of knots and the hyperbolic geometry of knot complements . Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.
7. The Whitehead Conjecture
Equation: G = (S | R)
- when CW complex K (S | R) is aspherical
- if π2 (K (S | R)) = 0
What you are doing in this equation is prove the claim made by Mr. Whitehead in 1941 in an algebraic topology that every subcomplex of an aspherical CW complex that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.
8. Equation Eight
Equation: (EQ4)
- Where Γ = a second countable locally compact group
- And the * and r subscript = 0 or 1.
This equation is the definition of morphism and is referred to as an assembly map. Check out the reduced C*-algebra for more insight into the concept surrounding this equation.
9. The Euler-Mascheroni Constant
Equation: y=limn→∞(∑m=1n1m−log(n))
Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts. The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.
This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.
10. Equation Ten
Equation: π + e
Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of algebraic real numbers and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.
As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.
See the 11 Comments below.
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Posted in Mathematics category - 10 Sep 2020 [ Permalink ]
11 Comments on “10 Math Equations That Have Never Been Solved”
But 2(2127)−1 = 340282366920938463463374607431768211455 is not a prime number. It is divisible by 64511.
Hello I am explorer and i type on google search " unsolvable mathematical formulas ", and I first find this syte. I see you are good-math-guys. Do you know what is this formula means:
π × ∞ = " 5 "
If you happen to have a quantum computer, I am not kidding be smart and don't insert this formula: [π × ∞ = " 5 "] into it please.
Maybe only, if you know meaning of this three symbols up writen and connected together.
(x dot epsilon)
I can explain my theory if you want me to spoil the pleasure of solving the equation. And mathematics as a science too or " as well " sorry i am not good in English, and google translate is not exelent.
8.539728478 is the answer to number 10
8.539728478 is the answer to number 10 or 8.539734221
Equation Four: Solved
To determine whether the number 2(2^127)-1 – 1 is a prime number, we first need to calculate its value. The expression 2(2^127) can be simplified as follows:
2(2^127) = 2 * 2^127 = 2^128
Therefore, the expression 2(2^127)-1 – 1 can be written as 2^128 – 1 – 1. We can then simplify this further to get:
2^128 – 1 – 1 = 2^128 – 2
To determine whether this number is prime, we can use the fundamental theorem of arithmetic, which states that every positive integer can be written as a product of prime numbers in a unique way (ignoring the order of the factors). This means that if a number is not prime, it can be expressed as the product of two or more prime numbers.
We can use this theorem to determine whether 2^128 – 2 is prime by trying to express it as the product of two or more prime numbers. However, it is not possible to do this, because 2^128 – 2 cannot be evenly divided by any prime number (except for 1, which is not considered a prime number).
Therefore, we can conclude that 2^128 – 2 is a prime number, because it cannot be expressed as the product of two or more prime numbers.
Equation Ten: Solved
The sum of π and e is equal to π + e = 3.14159 + 2.71828 = 5.85987.
To determine whether this number is algebraic or transcendental, we first need to understand the difference between these two types of numbers. Algebraic numbers are numbers that can be expressed as a root of a polynomial equation with integer coefficients, while transcendental numbers cannot be expressed in this way.
In this case, the number 5.85987 can be expressed as the root of the polynomial equation x^2 - 5.85987x + 2.71828 = 0. Therefore, it is an algebraic number.
In conclusion, the sum of π and e is equal to 5.85987, which is an algebraic number.
Equation 2: SOLVED
The equation 3n + 1 states that a positive integer n should be multiplied by 3 and then 1 should be added to the result. If the resulting value is then divided by 2 and the quotient is a non-negative integer, the process should be repeated with the new value of n.
To prove that this equation always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can start by substituting a value for n and performing the calculations as specified in the equation.
For example, if n is equal to 1, the sequence of values will be: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2 = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5
The value of n becomes 5 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 5 3n + 1 = 3(5) + 1 = 16 n = 16/2 = 8 n = 8/2 = 4 n = 4/2 = 2 n = 2/2 = 1 n = 1/2 = 0.5
Since n must be a non-negative integer, the value of n becomes 1 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2
To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.
If n is equal to 4, the sequence of values will be: n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5
Since n must be a non-negative integer, the value of n becomes 6 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 6 3n + 1 = 3(6) + 1 = 19 n = 19/2 = 9.5
Certainly! To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.
If n is equal to 4, the sequence of values will be:
n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5
Since n must be a non-negative integer, the value of n becomes 9 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 9 3n + 1 = 3(9) + 1 = 28 n = 28/2 = 14 n = 14/2 = 7 n = 7/2 = 3.5
The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5 n = 5/2 = 2.5
Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
As we can see, the sequence of values becomes repetitive
The Riemann Hypothesis
This equation states that the sum of the positive integers divisible by n (σ(n)) is less than or equal to the n-th harmonic number (Hn) plus the natural logarithm of the n-th harmonic number (ln(Hn)) multiplied by the n-th harmonic number (Hn) raised to the power of Hn.
To solve this equation, you would need to substitute a specific value for n and determine the value of Hn and σ(n) for that specific value. You can then substitute these values into the equation and see if it holds true.
For example, if n = 5, the sum of the positive integers divisible by 5 (σ(5)) is 15 (1 + 5 + 10 + 15 + 20 + 25), the 5th harmonic number (H5) is 2.28, and the natural logarithm of the 5th harmonic number (ln(H5)) is 0.83. Substituting these values into the equation, we get:
σ(5) ≤ H5 + ln(H5)eH5 15 ≤ 2.28 + 0.83 * 2.28^2.28 15 ≤ 4.39
Since 15 is less than or equal to 4.39, the equation holds true for this specific value of n.
Equation #9
In the equation y = limn→∞(∑m=1n1m−log(n)), y is the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity.
The Euler-Mascheroni constant is defined as the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity, and it has a value of approximately 0.5772. Therefore, y is equal to the Euler-Mascheroni constant, which is a rational number.
Rational numbers are numbers that can be expressed as the ratio of two integers, such as 3/4, 7/11, or 2/5. They can be written as a finite or repeating decimal, such as 0.75, 0.636363636..., or 1.5.
Irrational numbers are numbers that cannot be expressed as the ratio of two integers, and they cannot be written as a finite or repeating decimal. Examples of irrational numbers include √2, π, and e.
Since y is equal to the Euler-Mascheroni constant, which is a rational number, y is a rational number.
The equation G = (S | R) is a definition of a CW complex, where S and R are subcomplexes of G. A CW complex is a topological space that can be built up from cells, where each cell is homeomorphic to a closed ball in Euclidean space.
The statement "when CW complex K (S | R) is aspherical" means that the complex K (S | R) does not contain any non-trivial loops, i.e. loops that cannot be continuously contracted to a point. This implies that the fundamental group of K (S | R) is trivial, which means that π1(K (S | R)) = {e}.
The statement "if π2 (K (S | R)) = 0" means that the second homotopy group of the complex K (S | R) is trivial, which means that there are no non-trivial 2-dimensional holes in K (S | R).
Together, these statements imply that the CW complex K (S | R) is a topological space with no non-trivial loops or holes. This is a strong condition that is satisfied by very few spaces, and it is a necessary condition for a space to be aspherical.
In summary, the statement "when CW complex K (S | R) is aspherical" and "if π2 (K (S | R)) = 0" implies that the complex K (S | R) is a topological space with no non-trivial loops or holes, which is a necessary condition for a space to be aspherical.
#3 Erdos Strauss Conjecture:
To solve the equation 4/n = 1/a + 1/b + 1/c where n ≥ 2, a, b and c are positive integers, we can first multiply both sides of the equation by nabc to get rid of the fractions:
4abc = nab + nbc + nac
We can then group like terms:
4abc = (n + a)(b + c)
Now we can use the fact that n, a, b, and c are positive integers to make some observations:
Since n, a, b and c are positive integers, n, a, b and c must be factors of 4abc. Since n is greater than or equal to 2, it must be one of the factors of 4abc. The other factors of 4abc are (n + a), b, and c. So, to find all the possible values of n, a, b, and c, we must find all the ways to factorize 4abc such that one of the factors is greater than or equal to 2.
4abc = 4 * 1 * 1 * 2 * 3 * 5 = 120
Some possible factorizations are:
n = 2, a = 1, b = 5, c = 12 n = 2, a = 3, b = 5, c = 8 n = 2, a = 4, b = 3, c = 15 n = 2, a = 6, b = 2, c = 20 n = 4, a = 1, b = 3, c = 30 So, the possible solutions to the equation are: (n,a,b,c) = (2,1,5,12), (2,3,5,8), (2,4,3,15), (2,6,2,20), (4,1,3,30)
It's worth noting that this is not an exhaustive list, but just some of the possible solutions, as there could be infinitely many solutions to this equation.
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Home / Blog / Math / Hardest Math Questions That Are Surprisingly Easy To Solve!
Hardest Math Questions That Are Surprisingly Easy To Solve!
Have you ever struggled enough with a math problem only to realize that it was too easy to solve? Many students will definitely vouch for it—a math question looked quite simple when they sat with it, but within minutes they understood that they had no idea how to solve it!
Sometimes you come across some of the most difficult math questions that are both challenging and brain teasers. Even if you are well-prepared for your exam, you will encounter difficult math questions. But here’s the thing: They can be solved in a flash, regardless of how ridiculously difficult they appear.
Yes, you heard that right! If you understand basic math skills and concepts, all you need are some math tricks and practice to solve the problems in a jiffy.
Check out These 10 Hardest Math Questions That will Test Your Logic and Problem-solving Skills
Interestingly, these hard math problems can be solved easily with basic math operators. Ready to test your math skills?
- A bat and a ball cost one dollar and ten cents in total. The bat costs a dollar more than the ball. How much does the ball cost?
Answer: The ball costs 5 cents. How did that happen? If you were thinking of some other answers to this problem, then this explanation might help you. Well, to simplify the question, the difference between $1 and 10 cents is 90 cents, not $1. The only way for the bat to cost a dollar more than the ball while still having the total cost equal to $1.10 is for the baseball bat to cost $1.05 and the ball to cost 5 cents.
- The classic PEMDAS problem: 6 ÷ 2(1+2) = ?
Answer: 9 Every PEMDAS problem leads to a huge disruption of opinions, splitting the masses into various answers while everyone believes that they have the correct answer. So, how was this solved and how did they arrive at “9” as the solution? If you remember the order of operations of PEMDAS from grade school, one generally works through the problem by solving parentheses, then the exponents, multiplication, and division, followed by addition and subtraction. Sometimes, some people interpret PEMDAS differently, and that is when the disagreement begins. According to the PEMDAS rule, one should solve anything inside parentheses, then exponents, and then all multiplication and division, starting from the left to the right, as per the operations.
- Replace the question mark in the above diagram with an appropriate number.
Answer: 6 If you enjoy playing Sudoku, then this hard math problem would have been quite a breeze for you! The rows and columns all add up to 15, and that is how the answer turned out to be 6.
- Solve the unfinished equation:
Answer: 4 = 256 Have a close look at the equation again, and find out the common thing among the equations. There is a formula that has been used for these equations. 4^x = Y. So, 4^1 = 4, 4^2 = 16, 4^3 = 64, and 4^4 = 256.
- The Lily Pad problem: A lake has a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long will it take for the patch to cover half of the lake?
Answer: It will take 47 days for the patch to cover half of the lake. If you assumed that half of the lake would be covered in half of the time, then your assumption could be incorrect. Here, in the math problem, it says that the patch “doubles” in size, which means, given any day, the lily pad was half the size the day before. So if the patch covers up the entire lake on the 48th day, it means the lily pad was half the size of the lake on day 47.
- When you add five to nine, you get two. The answer is correct, but how?
Answer: Well, here’s the trick. When it is 9 a.m., you add five hours to it. Voila, you get 2 p.m. as your answer! Getting smarter every minute, aren’t we?
- A group of students was standing in the sun facing due west on a march past event. The captain shouted at them: Right turn! About turn! Left turn! At the end of these commands, which direction are the students facing now?
Answer: East. Well, this one might seem tricky, but the answer is quite simple. The students will first turn 90 degrees in a right turn and then 180 degrees in an about-turn. The students will finally turn 90 degrees in a left turn. Therefore, the students are now facing east.
- Guess how many triangles are there in the above picture?
Answer: 18 This is another tricky math question where different people have come up with different answers. While some forget to count the hidden triangles, others forget to count the giant triangle. Why don’t you concentrate on the triangle again and find all the triangles?
- How to make the following equation precise using three of these four mathematical symbols: + – × ÷ without following any order as such?
Equation: 21 _ 3 _ 18 _ 6 = 6
Answer: 21 – 3 + 18 ÷ 6 = 6 This is more of a symbol math problem than a PEMDAS problem. You can keep testing with the math operators until you find the correct answer.
- John was asked to paint numbers outside 100 apartments. This means he will have to paint numbers one through 100. How many times will he have to paint the number eight?
Answer: 20 times If you carefully read the question, the answer is within the question itself. It’s a tricky one, but this is how it goes: John paints on apartment 8, then he paints on apartment 18 (where the 8 digit comes up again) and, similarly, 28th, 38th, 48th, 58th, 68th, 78th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, and 98th apartments.
These were some of the hardest math questions, which probably left your mind boggled. You can always keep improving your logical and math skills with these kinds of math problems that might seem difficult but are easier than you had thought. So, keep testing yourself with more tricky math problems and get your logical juices flowing.
To read more about versatile and new online techniques to improve your math concepts and boost your confidence in solving difficult problems, visit Byju’s FutureSchool blog.
References:
- Jamie Spafford on Twitter: “Here’s a riddle for you: If a baseball bat and a baseball together cost $1.10 and the bat costs a dollar more than the ball, how much does the ball cost? (H/T @ScottMonty for including it in his really great article: https://t.co/foD6RdOUgy)” / Twitter . (n.d.). Retrieved April 22, 2022, from https://twitter.com/jamiespafford/status/1290641293891317770
- 20 Tricky But Fun Grade-School Math Questions – Hard Math Problems . (n.d.). Retrieved April 22, 2022, from https://bestlifeonline.com/tricky-math-questions/
- 34 Hard Math Riddles and Word Problems for Future Geniuses | Fatherly . (n.d.). Retrieved April 22, 2022, from https://www.fatherly.com/play/hard-math-riddles-for-kids-with-answers/
- Southern Math Riddles Jeopardy . (n.d.). Retrieved April 22, 2022, from https://jeopardylabs.com/print/math-riddles-jeopardy
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Sat / act prep online guides and tips, the hardest act math question types.
You’ve studied and now you’re geared up for the ACT math section (whoo!). But are you ready to take on the most challenging math questions the ACT has to offer?
If you're up for the challenge, preparation starts by understanding the toughest question types you'll face on the ACT . So if you’ve got your heart set on that perfect score (or you’re just really curious to see what the most difficult questions will be), then this is the guide for you.
We’ve put together what we believe to be the most most difficult math question types the ACT has given to students in the past 10 years, complete with strategies and answer explanations for each . These are all real ACT math questions, so understanding and studying them is one of the best ways to improve your current ACT score and knock it out of the park on test day.
Let's dive in!
Brief Overview of the ACT Math Section
Like all topic sections on the ACT, the ACT math section is one complete section that you will take all at once. It will always be the second section on the test and you will have 60 minutes to completed 60 questions .
The ACT arranges its questions in order of ascending difficulty. As a general rule of thumb, questions 1-20 will be considered “easy,” questions 21-40 will be considered “medium-difficulty,” and questions 41-60 will be considered “difficult.”
The way the ACT classifies “easy” and “difficult” is by how long it takes the average student to solve a problem as well as the percentage of students who answer the question correctly. The faster and more accurately the average student solves a problem, the “easier” it is. The longer it takes to solve a problem and the fewer people who answer it correctly, the more “difficult” the problem.
(Note: we put the words “easy” and “difficult” in quotes for a reason—everyone has different areas of math strength and weakness, so not everyone will consider an “easy” question easy or a “difficult” question difficult. These categories are averaged across many students for a reason and not every student will fit into this exact mold.)
All that being said, with very few exceptions, the most difficult ACT math problems will be clustered in the far end of the test. Besides just their placement on the test, these questions share a few other commonalities. We'll take a look at example questions and how to solve them and at what these types of questions have in common, in just a moment.
But First: Should You Be Focusing on the Hardest Math Questions Right Now?
If you’re just getting started in your study prep, definitely stop and make some time to take a full practice test to gauge your current score level and percentile. The absolute best way to assess your current level is to simply take the ACT as if it were real , keeping strict timing and working straight through (we know—not the most thrilling way to spend four hours, but it will help tremendously in the long run). So print off one of the free ACT practice tests available online and then sit down to take it all at once.
Once you’ve got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate ACT score. If you’re currently scoring in the 0-16 or 17-24 range , your best best is to first check out our guides on using the key math strategies of plugging in numbers and plugging in answers to help get your score up to where you want it to. Only once you've practiced and successfully improved your scores on questions 1-40 should you start in trying to tackle the most difficult math problems on the test.
If, however, you are already scoring a 25 or above and want to test your mettle for the real ACT, then definitely proceed to the rest of this guide . If you’re aiming for perfect (or close to) , then you’ll need to know what the most difficult ACT math questions look like and how to solve them. And luckily, that’s exactly what we’re here for.
Hardest Types of ACT Math Questions
Now that you're positive that you should be trying out these difficult math questions, let’s get right to it! We've broken hard ACT questions into sections based on type. Additionally, the answers to these questions are in a separate section below, so you can go through them all at once without getting spoiled.
Type #1: Graph Interpretation
Knowing how to read and interpret graphs can be tricky, especially since that information can also require you to use other math skills, like finding points on a line or calculating angles.
To unpack a graph question, first ask yourself—what is the prompt actually asking me to do? Figuring out the core mathematical concept behind the graphed information will make it easier to answer the question.
Check out the two graph interpretation questions below. Notice that both are asking you to interpret the graph data by using other mathematical concepts! With that in mind, try solving them on your own to get practice solving these tough questions. (Remember: we've included the answers below!)
Graph Interpretation Questions
QUESTION #1:
In the figure below, line $q$ in the standard $(x,y)$ coordinate plane has equation $-2x+y=1$ and intersects line $r$, which is distinct from line $q$, at a point on the $x$-axis. The angles $∠a$ and $∠b$ formed by these lines and the $x$-axis are congruent. What is the slope of line $r$?
G. $-{1}/2$
K. Cannot be determined
First, turn our given equation for line q into proper slope-intercept form .
Now, we are told that the angles the lines form are congruent. This means that the slopes of the lines will be opposites of one another [Note: perpendicular lines have opposite reciprocal slopes, so do NOT get these concepts confused!].
Since we have already established that the slope of line $q$ is 2, line $r$ must have a slope of -2.
Our final answer is F , -2
QUESTION #2:
The graph of the equation $h=-at^2 + bt + c$, which describes how the height, $h$, of a hit baseball changes over time, $t$, is shown below.
If you alter only this equation's $c$ term, which gives the height at time $t=0$, the alteration has an effect on which of the following?
I. The $h$-intercept II. The maximum value of $h$ III. The $t$-intercept
H. III only
J. I and III only
K. I, II, and III
The equation we are given ($−at^2+bt+c$) is a parabola and we are told to describe what happens when we change $c$ (the $y$-intercept).
From what we know about functions and function translations , we know that changing the value of $c$ will shift the entire parabola upwards or downwards, which will change not only the $y$-intercept (in this case called the "$h$-intercept"), but also the maximum height of the parabola as well as its $x$-intercept (in this case called the $t$-intercept). You can see this in action when we raise the value of the $y$-intercept of our parabola.
Options I, II, and III are all correct.
Our final answer is K, I, II, and III
Type #2: Trigonometry Questions
Trigonometry questions are tough for a number of reasons. First, they require you to memorize formulas and values in order to properly interpret questions. There's also a lot of conversion involved: you have to know how (and when!) to convert radians to degrees and visa versa.
And don't forget functions! Trigonometry uses non-linear functions, which is tough because its graph isn't a line or part of a line. Because trigonometry combines so many advanced math concepts, they can trip you up.
Test your trig skills on the advanced questions below. Try to work the, without peeking at the answers!
Trigonometry Questions
The equations of the two graphs shown below are $y_1(t) = a_1\sin (b_1t)$ and $y_2(t)=a_2\cos(b_2t)$, where the constants $b_1$ and $b_2$ are both positive real numbers.
Which of the following statements is true of the constants $a_1$ and $a_2$?
A. $0 < a_1 < a_2$
B. $0 < a_2 < a_1$
C. $a_1 < 0 < a_2$
D. $a_1 < a_2 < 0$
E. $a_2 < a_1 < 0$
The position of the a values (in front of the sine and cosine) means that they determine the amplitude (height) of the graphs. The larger the a value, the taller the amplitude.
Since each graph has a height larger than 0, we can eliminate answer choices C, D, and E.
Because $y_1$ is taller than $y_2$, it means that $y_1$ will have the larger amplitude. The $y_1$ graph has an amplitude of $a_1$ and the $y_2$ graph has an amplitude of $a_2$, which means that $a_1$ will be larger than $a_2$.
Our final answer is B , $0 < a_2 < a_1$.
For x such that $0 < x <π/2$, the expression ${√{1 – \cos^2 x}/{\sin x} + {√{1 – \sin^2 x}/{\cos x}$ is equivalent to:
J. $-\tan{x}$
K. $\sin{2x}$
If you remember your trigonometry shortcuts , you know that $1−{\cos^2}x+{\cos^2}x=1$. This means, then, that ${\sin^2}x=1−{\cos^2}x$ (and that ${\cos^2}x=1−{\sin^2}x$).
So we can replace our $1−{\cos^2}x$ in our first numerator with ${\sin^2}x$. We can also replace our $1−{\sin^2}x$ in our second numerator with ${\cos^2}x$. Now our expression will look like this:
${√{\sin^2 x}/{\sin x}+{√{\cos^2 x}/{\cos x }$
We also know that the square root of a value squared will cancel out to be the original value alone (for example,$√{2^2}=2$), so our expression will end up as:
$={\sin{x}}/{\sin{x}}+{\cos{x}}/{\cos{x}}$
Or, in other words:
Our final answer is H , 2.
Type #3: Logarithms
Logarithms can be hard because they require you to differentiate between exponential and logarithmic questions and understand logarithmic notation. If you can't read the question, you can't answer it, after all!
You'll also need to understand exponents to work with logarithms since the two concepts are intertwined. Working through the following equation can help you learn how to manage these tough problems.
Logarithm Questions
What is the real value of $x$ in the equation $\log_2{24} – \log_2{3} = \log_5{x}$?
ANSWER #1: If you’ve brushed up on your log basics, you know that $\log_b(m/n)=\log_b(m)−\log_b(n)$. This means that we can work this backwards and convert our first expression into:
$\log_2(24)-\log_2(3)=\log_2(24/3)$
$=\log_2(8)$
We also know that a log is essentially asking: "To what power does the base need to raised in order to achieve this certain value?" In this particular case, we are asking: "To which power must 2 be raised to equal 8?" The answer to this question is 3, since $(2^3=8)$, so $\log_2(8)=3$
Now this expression is equal to $\log_5(x)$, which means that we must also raise our 5 to the power of 3 in order to achieve $x$. So:
$3=\log_5(x)$
Our final answer is J , 125.
Type #4: Functions
Functions are one of those question types that are tough for some students but easier for others. That's because as functions get more complex, so does finding the correct answer. If you understand how functions work, you can break down more complex problems into simpler steps.;
Try your hand at the hard function question below to test your knowledge.
Function Questions
The functions $y = \sin x \and y = \sin(x + a) + b$, for constants $a$ and $b$, are graphed in the standard $(x,y)$ coordinate plane below. The functions have the same maximum value. One of the following statements about the values of $a$ and $b$ is true. Which statement is it?
A. $a < 0$ and $b = 0$
B. $a < 0$ and $b > 0$
C. $a = 0$ and $b > 0$
D. $a > 0$ and $b< 0$
E. $a > 0$ and $b> 0$
The only difference between our function graphs is a horizontal shift , which means that our b value (which would determine the vertical shift of a sine graph) must be 0.
Just by using this information, we can eliminate every answer choice but A, as that is the only answer with $b=0$. For expediency's sake, we can stop here.
Our final answer is A , $a<0$ and $b=0$
Advanced ACT Math note : An important word in ACT Math questions is "must," as in "something must be true." If a question doesn't have this word, then the answer only has to be true for a particular instance (that is, it could be true.)
In this case, the majority of the time, for a graph to shift horizontally to the left requires $a>0$. However, because $\sin(x)$ is a periodic graph, $\sin(x+a)$ would shift horizontally to the left if $a=-π/2$, which means that for at least one value of the constant $a$ where $a<0$, answer A is true. In contrast, there are no circumstances under which the graphs could have the same maximum value (as stated in the question text) but have the constant $b≠0$.
As we state above, though, on the real ACT, once you reach the conclusion that $b=0$ and note that only one answer choice has that as part of it, you should stop there. Don't get distracted into wasting more time on this question by the bait of $a<0$!
Whoo! You made it to the finish line—go you!
What Do the Hardest ACT Math Questions Have in Common?
Now, lastly, before we get to the questions themselves, it is important to understand what makes these hard questions “hard.” By doing so, you will be able to both understand and solve similar questions when you see them on test day, as well as have a better strategy for identifying and correcting your previous ACT math errors.
In this section, we will look at what these questions have in common and give examples for each type. In the next section, we will give you all 21 of the most difficult questions as well as answer explanations for each question, including the ones we use as examples here.
Some of the reasons why the hardest math questions are the hardest math questions are because the questions do the following:
#1: Test Several Mathematical Concepts at Once
EXAMPLE QUESTION:
Consider the functions $f(x) = √{x}$ and $g(x) = 7x + b$. In the standard $(x,y)$ coordinate plane, $y = f(g(x))$ passes through $(4,6)$. What is the value of $b$?
E. $4 – {7√6}$
As you can see, this question deals with a combination of functions and coordinate geometry points.
#2: Require Multiple Steps
Many of the most difficult ACT Math questions primarily test just one basic mathematical concept. What makes them difficult is that you have to work through multiple steps in order to solve the problem. (Remember: the more steps you need to take, the easier it is to mess up somewhere along the line!)
An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit?
A. $19/900$
B. $81/900$
C. $90/900$
D. $171/900$
E. $271/{1{,}000}$
Though it may sound like a simple probability question, you must run through a long list of numbers with 0 as a digit. This leaves room for calculation errors along the way.
#3: Use Concepts You're Less Familiar With
Another reason the questions we picked are so difficult for many students is that they focus on subjects you likely have limited familiarity with. For example, many students are less familiar with algebraic and/or trigonometric functions than they are with fractions and percentages, so most function questions are considered “high difficulty” problems.
The functions $y =$ sin$x$ and $y =$ sin$(x + a) + b$, for constants $a$ and $b$, are graphed in the standard $(x,y)$ coordinate plane below. The functions have the same maximum value. One of the following statements about the values of $a$ and $b$ is true. Which statement is it?
D. $a > 0$ and $b < 0$
E. $a > 0$ and $b > 0$
Many students get intimidated with function problems because they lack familiarity with these types of questions.
#4: Give You Convoluted or Wordy Scenarios to Work Through
Some of the most difficult ACT questions are not so much mathematically difficult as they are simply tough to decode. Especially as you near the end of the math section, it can be easy to get tired and misread or misunderstand exactly what the question is even asking you to find.
In the complex plane, the horizontal axis is called the real axis and the vertical axis is called the imaginary axis. The complex number $a + bi$ graphed in the complex plane is comparable to the point $(a,b)$ graphed in the standard $(x,y)$ coordinate plane.
The modulus of the complex number $a + bi$ is given by ${√{a^2+b^2}$ . Which of the complex numbers $z_1$ , $z_2$ , $z_3$ , $z_4$ , and $z_5$ below has the greatest modulus?
This question presents students with a completely foreign mathematical concept and can eat up the limited available time.
#5: Appear Deceptively Easy
Remember—if a question is located at the very end of the math section, it means that a lot of students will likely make mistakes on it. Look out for these questions, which may give a false appearance of being easy in order to lure you into falling for bait answers. Be careful!
Which of the following number line graphs shows the solution set to the inequality $|x–5|<1$ ?
This question may seem easy, but, because of how it is presented, many students will fall for one of the bait answers.
#6: Involve Multiple Variables or Hypotheticals
The more difficult ACT Math questions tend to use many different variables—both in the question and in the answer choices—or present hypotheticals. (Note: The best way to solve these types of questions—questions that use multiple integers in both the problem and in the answer choices—is to use the strategy of plugging in numbers .)
If $x$ and $y$ are real numbers such that $x > 1$ and $y < -1$, then which of the following inequalities must be true?
A. ${x/y} > 1$
B. ${|x|^2} > |y|$
C. ${x/3} – 5 > {y/3} – 5$
D. ${x^2} + 1 > {y^2} + 1$
E. ${x^{-2}} > {y^{-2}}$
Working with hypothetical scenarios and variables is almost always more challenging than working with numbers.
The Take-Aways
Taking the ACT is a long journey; the more you get acclimated to it ahead of time, the better you'll feel on test day. And knowing how to handle the hardest questions the test-makers have ever given will make taking your ACT seem a lot less daunting.
If you felt that these questions were easy , make sure not underestimate the effect of adrenaline and fatigue on your ability to solve your math problems. As you study, try to follow the timing guidelines (an average of one minute per ACT math question) and try to take full tests whenever possible. This is the best way to recreate the actual testing environment so that you can prepare for the real deal.
If you felt these questions were challenging , be sure to strengthen your math knowledge by checking out our individual math topic guides for the ACT . There, you'll see more detailed explanations of the topics in question as well as more detailed answer breakdowns.
What’s Next?
Felt that these questions were harder than you were expecting? Take a look at all the topics covered on the ACT math section and then note which sections you had particular difficulty in. Next, take a look at our individual math guides to help you strengthen any of those weak areas.
Running out of time on the ACT math section? Our guide to helping you beat the clock will help you finish those math questions on time.
Aiming for a perfect score? Check out our guide on how to get a perfect 36 on the ACT math section , written by a perfect-scorer.
Want to improve your ACT score by 4 points?
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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.
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Hardest math problem ever solved
There are no whole number solutions to the equation xn + yn = zn when n is greater than 2. Otherwise known as Fermat's Last Theorem, this equation was first posed by French mathematician Pierre de Fermat in 1637, and had stumped the world's brightest minds for more than 300 years.Mar 17, 2016
This Is The Hardest Math Problem In The World
The Riemann hypothesis is one of the Millenium Prize Problems, a list of unsolved math problems compiled by the Clay Institute. The Clay Institute has offered a
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The Most Difficult Math Problem In The History! (358 Years To
Can you solve the hardest maths problems ever?
Many difficult math problems have been solved throughout history, but one of the most famous and challenging problems is considered to be Fermat's Last Theorem.
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These Are the 10 Hardest Math Problems Ever Solved
But those itching for their Good Will Hunting moment, the Guinness Book of Records puts Goldbach's Conjecture as the current longest-standing
Math is a challenging subject for many students, but with practice and persistence, anyone can learn to figure out complex equations.
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… The 10 Hardest Math Problems That Remain Unsolved …
The smartest people in the world can’t crack them. maybe you’ll have better luck..
For all the recent strides we’ve made in the math world, like how a supercomputer finally solved the Sum of Three Cubes problem that puzzled mathematicians for 65
years, we’re forever crunching calculations in pursuit of deeper numerical knowledge.
Some math problems have been challenging us for centuries, and while brain-busters like the ones that follow may seem impossible, someone is bound to solve ‘em
eventually.
For now, take a crack at the toughest math problems known to man, woman, and machine.
Earlier this month, news broke of progress on this 82-year-old question, thanks to prolific mathematician Terence Tao.
And while the story of Tao’s breakthrough is good news, the problem isn’t fully solved.
A refresher on the Collatz Conjecture: It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled
and then added to 1.
Take any natural number, apply f, then apply f again and again.
You eventually land on 1, for every number we’ve ever checked.
The Conjecture is that this is true for all natural numbers.
Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways.
But his methods most likely can’t be adapted to yield a complete solution to the problem, as he subsequently explained.
So we might be working on it for decades longer.
The Conjecture is in the math discipline known as Dynamical Systems, or the study of situations that change over time in semi-predictable ways.
It looks like a simple, innocuous question, but that’s what makes it special.
Why is such a basic question so hard to answer?
It serves as a benchmark for our understanding; once we solve it, then we can proceed to much more complicated matters.
The study of dynamical systems could become more robust than anyone today could imagine.
But we’ll need to solve the Collatz Conjecture for the subject to flourish.
One of the biggest unsolved mysteries in math is also very easy to write.
Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes.”
You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19.
Computers have checked the Conjecture for numbers up to some magnitude.
But we need proof for all natural numbers.
Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler,
considered one of the greatest in math history.
As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”
Euler may have sensed what makes this problem counterintuitively hard to solve.
When you look at larger numbers, they have more ways of being written as sums of primes, not less.
Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23.
So it feels like Goldbach’s Conjecture is an understatement for very large numbers.
Still, a proof of the conjecture for all numbers eludes mathematicians to this day.
It stands as one of the oldest open questions in all of math.
Together with Goldbach’s, the Twin Prime Conjecture is the most famous in the subject of math called Number Theory, or the study of natural numbers and their
properties, frequently involving prime numbers.
Since you’ve known these numbers since grade school, stating the conjectures is easy.
When two primes have a difference of 2, they’re called twin primes.
So 11 and 13 are twin primes, as are 599 and 601.
Now, it’s a Day 1 Number Theory fact that there are infinitely many prime numbers.
So, are there infinitely many twin primes?
The Twin Prime Conjecture says yes.
Let’s go a bit deeper.
The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6.
And so the second twin prime is always 1 more than a multiple of 6.
You can understand why, if you’re ready to follow a bit of Number Theory.
All primes after 2 are odd.
Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6.
Well, one of those three possibilities for odd numbers causes an issue.
If a number is 3 more than a multiple of 6, then it has a factor of 3.
Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself).
And that’s why every third odd number can’t be prime.
How’s your head after that paragraph?
Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.
The good news is we’ve made some promising progress in the last decade.
Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture.
This was their idea: Trouble proving there are infinitely many primes with a difference of 2?
How about proving there are infinitely many primes with a difference of 70,000,000.
That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.
For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds.
Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture.
The closest we’ve come —given some subtle technical assumptions—is 6.
Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.
Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math.
It’s one of the seven Millennium Prize Problems , with a million dollar reward for its solution.
It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.
There is a function, called the Riemann zeta function, written in the image above.
For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s.
For example, if s=2, then 𝜁(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6.
When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.
So tricky, in fact, that it’s become the ultimate math question.
Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.”
On the plane of complex numbers, this means the function has a certain behavior along a special vertical line.
You can see this in the visualization of the function above—it’s along the boundary of the rainbow and the red.
The hypothesis is that the behavior continues along that line infinitely.
The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.
If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.
The Birch and Swinnerton-Dyer Conjecture is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain
This Conjecture involves the math topic known as Elliptic Curves.
When we recently wrote about the toughest math problems that have been solved , we mentioned one of the greatest achievements in 20th century math: the solution
to Fermat’s Last Theorem.
It was solved by Sir Andrew Wiles, using Elliptic Curves.
So you could call this a very powerful new branch of math.
In a nutshell, an elliptic curve is a special kind of function.
They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and
Number Theory.
British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s.
Its exact statement is very technical, and has evolved over the years.
One of the main stewards of this evolution has been none other than Wiles.
To see its current status and complexity, check out this famous update by Wells in 2006.
A broad category of problems in math are called the Sphere Packing Problems.
They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space,
like fruit at the grocery store.
Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.
When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6
neighboring spheres, then your kissing number is 6.
Nothing tricky.
A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation.
But a basic question about the kissing number stands unanswered.
First, a note on dimensions.
Dimensions have a specific meaning in math: they’re independent coordinate axes.
The x-axis and y-axis show the two dimensions of a coordinate plane.
When a character in a sci-fi show says they’re going to a different dimension, that doesn’t make mathematical sense.
You can’t go to the x-axis.
A 1-dimensional thing is a line, and 2-dimensional thing is a plane.
For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions.
It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right.
There’s proof of an exact number for 3 dimensions, although that took until the 1950s.
Beyond 3 dimensions, the Kissing Problem is mostly unsolved.
Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see on this chart .
For larger numbers, or a general form, the problem is wide open.
There are several hurdles to a full solution, including computational limitations.
So expect incremental progress on this problem for years to come.
The simplest version of the Unknotting Problem has been solved, so there’s already some success with this story.
Solving the full version of the problem will be an even bigger triumph.
You probably haven’t heard of the math subject Knot Theory.
It’s taught in virtually no high schools, and few colleges.
The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.
For example, you might know how to tie a “square knot” and a “granny knot.”
They have the same steps except that one twist is reversed from the square knot to the granny knot.
But can you prove that those knots are different?
Well, knot theorists can.
Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing.
The cool news is that this has been accomplished!
Several computer algorithms for this have been written in the last 20 years, and some of them even animate the process .
Where the Unknotting Problem remains is computational.
In technical terms, it’s known that the Unknotting Problem is in NP, while we don’t know if it’s in P.
That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an
impossibly long time.
If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest.
On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound.
Eventually, we’ll find out.
If you’ve never heard of Large Cardinals , get ready to learn.
In the late 19th century, a German mathematician named Georg Cantor figured out that infinity comes in different sizes.
Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.
There is the first infinite size, the smallest infinity , which gets denoted ℵ₀.
That’s a Hebrew letter aleph; it reads as “aleph-zero.”
It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.
Next, some common sets are larger than size ℵ₀.
The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀.
But the reals aren’t that big; we’re just getting started on the infinite sizes.
For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals.
It ’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.”
Then, if their proof is good, that’s the new largest known cardinal.
Until someone else comes up with a larger one.
Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward.
There’s now even a beautiful wiki of known large cardinals , named in honor of Cantor.
So, will this ever end?
The answer is broadly yes, although it gets very complicated.
In some senses, the top of the large cardinal hierarchy is in sight.
Some theorems have been proven which impose a sort of ceiling on the possibilities for large cardinals.
But many open questions remain, and new cardinals have been nailed down as recently as 2019.
It’s very possible we will be discovering more for decades to come.
Hopefully we’ll eventually have a comprehensive list of all large cardinals.
Given everything we know about two of math’s most famous constants, 𝜋 and e, it’s a bit surprising how lost we are when they’re added together.
This mystery is all about algebraic real numbers .
The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients.
For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers.
The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.
All rational numbers, and roots of rational numbers, are algebraic.
So it might feel like “most” real numbers are algebraic.
Turns out it’s actually the opposite.
The antonym to algebraic is transcendental, and it turns out almost all real numbers are transcendental—for certain mathematical meanings of “almost all.”
So who’s algebraic, and who’s transcendental?
The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century.
You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?
Well, we do know that both 𝜋 and e are transcendental.
But somehow it’s unknown whether 𝜋+e is algebraic or transcendental.
Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them.
So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.
Here’s another problem that’s very easy to write, but hard to solve.
All you need to recall is the definition of rational numbers.
Rational numbers can be written in the form p/q, where p and q are integers.
So 42 and -11/3 are rational, while 𝜋 and √2 are not.
It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?
Meet the Euler-Mascheroni constant 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly;
it looks like the image above.
The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.”
So it’s a combination of two very well-understood mathematical objects.
It has other neat closed forms, and appears in hundreds of formulas.
But somehow, we don’t even know if 𝛾 is rational.
We’ve calculated it to half a trillion digits, yet nobody can prove if it’s rational or not.
The popular prediction is that 𝛾 is irrational.
Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it.
Sep 26, 2019
………. so park your political correctness for a bit……A smart Indian had three wives who all lived in the same teepee … so no privacy no nothing…. but like I said he’s a smart Indian……..decides to build each squaw their own teepee…. …first teepee he made out of rawhide….the second teepee he made out of deer hide and the third teepee…. went on the internet and got hippopotamus hide……….worked like you might imagine……9 months later the squaw in the rawhide teepee had her baby and shortly after that the squaw in the deer hide teepee had her baby and shortly after that the squaw in the hippopotamus hide teepee had twins……………………………..so you know what that proves?……………………….that the squaw of the hippopotamus is equal to the sum of the squaws of the other two hides…………. …………………..w
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The Legend of Question Six: One of The Hardest Maths Problems Ever
It's a secret to no one that maths is hard , so when you start talking about the hardest maths problems ever, things start to get a little crazy. Take the innocuously named Question 6, which is so complex, it can bring mathematicians to tears.
As mathematician Simon Pampena explains the Numberphile video above , the Legend of Question 6 spawned from a maths competition for high-schoolers held in Australia in 1988. (Yep, they make 'em tough down here.)
The competition was the International Mathematical Olympiad, which is held every year in a different country, and only six kids from every country are selected to compete. Points are scored on how each 'mathlete' performs on six different questions.
In 1988, the Australian Olympiad officials decided to throw a massive curveball to the kids on the final day of competition, and it's gone down in history as one of the toughest problems out there.
Just to give you an idea of how tough it was, Australian-American mathematician Terence Tao - recipient of the 2006 Fields Medal (the mathematician's 'Nobel Prize') - scored a 1 out of 7 when he attempted it. But, you know, he was 13 at the time, so let's cut the man some slack.
What made Question 6 so hard is that it actually tried to pay mind games with you as you solved it.
"In a way, they're actually designed to kind of throw you off if you know high school maths too well," says Pampena . "So say if you really know how to solve quadratic equations, and you see something that looks like a quadratic equation, it almost throws you down the wrong path."
Question 6 was actually submitted to the Australian Olympiad officials by a mathematician from West Germany, and the officials gave themselves SIX HOURS to solve it to see if it should be included in the event.
Not one official could solve Question 6 within the time limit. Some of the best mathematicians in the world at the time.
But they put it on a test for kids anyway, and only gave them about 90 minutes to solve it, because mathematicians are ridiculous.
So you just want to know wtf this problem is, right? Okay, here it is:
Let a and b be positive integers such that ab + 1 divides a 2 + b 2 . Show that a 2 + b 2 / ab + 1 is the square of an integer.
What's the solution? Well, we're going to let the Numberphile video above explain that one to you… nope sorry, there's no way they're fitting all that into a single video.
The video above will tell you what the question actually means, and then the video below will give you the answer:
Just please don't pop a vein in your head or something if you attempt it yourself. And if you somehow crush it, here's six more to try .
Home » Indiana University Of Pennsylvania » What Is The Hardest Math Ever?
What Is The Hardest Math Ever?
Table of Contents
These Are the 10 Toughest Math Problems Ever Solved
- The Collatz Conjecture. Dave Linkletter.
- Goldbach’s Conjecture Creative Commons.
- The Twin Prime Conjecture.
- The Riemann Hypothesis.
- The Birch and Swinnerton-Dyer Conjecture.
- The Kissing Number Problem.
- The Unknotting Problem.
- The Large Cardinal Project.
What is the hardest math thing to learn?
1. Algebra : Algebra is a branch of mathematics that studies symbols and the rules that control how they are used.
What branch of math is the hardest?
Algebra Algebra is the hardest branch of Maths. Abstract algebra particularly is the most difficult portion as it includes complex and infinite spaces.
What is the hardest math equation of all time?
For decades, a math puzzle has stumped the smartest mathematicians in the world. x 3 +y 3 +z 3 =k , with k being all the numbers from one to 100, is a Diophantine equation that’s sometimes known as “summing of three cubes.”
What are the 7 hardest math problems?
Clay “to increase and disseminate mathematical knowledge.” The seven problems, which were announced in 2000, are the Riemann hypothesis, P versus NP problem, Birch and Swinnerton-Dyer conjecture, Hodge conjecture, Navier-Stokes equation, Yang-Mills theory, and Poincaré conjecture .
Who created math?
Archimedes is known as the Father of Mathematics . Mathematics is one of the ancient sciences developed in time immemorial. A major topic of discussion regarding this particular field of science is about who is the father of mathematics.
What is the highest level of math?
Though Math 55 bore the official title “Honors Advanced Calculus and Linear Algebra,” advanced topics in complex analysis, point set topology, group theory, and differential geometry could be covered in depth at the discretion of the instructor, in addition to single and multivariable real analysis as well as abstract
What is the easiest math subject?
Which math classes are the easiest? According to a large group of high-schoolers, the easiest math class is Algebra 1 . That is the reason why most of the students in their freshman year end up taking Algebra 1. Following Algebra 1, Geometry is the second easiest math course in high school.
Is algebra or geometry harder?
Geometry is easier than algebra . Algebra is more focused on equations while the things covered in Geometry really just have to do with finding the length of shapes and the measure of angles.
Is algebra harder than calculus?
Linear algebra is easier than elementary calculus . In Calculus, you can get by without understanding the intuition behind theorems and just memorizing algorithms, which won’t work well in the case of linear algebra. By understanding the theorems in linear algebra, all questions can be solved.
What is the impossible math problem?
The Collatz conjecture is one of the most famous unsolved mathematical problems, because it’s so simple, you can explain it to a primary-school-aged kid, and they’ll probably be intrigued enough to try and find the answer for themselves.
What is this pi?
Succinctly, pi—which is written as the Greek letter for p, or π—is the ratio of the circumference of any circle to the diameter of that circle . Regardless of the circle’s size, this ratio will always equal pi. In decimal form, the value of pi is approximately 3.14.
What’s the hardest math question on earth?
What is the 1 million dollar math problem.
The Riemann hypothesis – an unsolved problem in pure mathematics, the solution of which would have major implications in number theory and encryption – is one of the seven $1 million Millennium Prize Problems. First proposed by Bernhard Riemann in 1859, the hypothesis relates to the distribution of prime numbers.
Can P vs NP be solved?
Although one-way functions have never been formally proven to exist, most mathematicians believe that they do, and a proof of their existence would be a much stronger statement than P ≠ NP. Thus it is unlikely that natural proofs alone can resolve P = NP .
What is the hardest math in high school?
What is the Hardest Math Class in High School? In most cases, you’ll find that AP Calculus BC or IB Math HL is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.
Who invented 0?
Brahmagupta “Zero and its operation are first defined by [Hindu astronomer and mathematician] Brahmagupta in 628,” said Gobets. He developed a symbol for zero: a dot underneath numbers.
Who invented pi?
The first calculation of π was done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians of the ancient world.
Is learning math hard?
Math seems difficult because it takes time and energy . Many people don’t experience sufficient time to “get” math lessons, and they fall behind as the teacher moves on. Many move on to study more complex concepts with a shaky foundation. We often end up with a weak structure that is doomed to collapse at some point.
Who passed Math 55?
Bill Gates Bill Gates took Math 55. To get a sense of the kind of brains it takes to get through Math 55, consider that Bill Gates himself was a student in the course. (He passed.) And if you’d like to sharpen your brain like Microsoft’s co-founder, here are The 5 Books Bill Gates Says You Should Read.
How many math levels are there?
The four math classes, from least challenging to most challenging are: Mathematical studies SL. Mathematics SL. Mathematics HL.
By Cary Hardy
Hi there! I'm Cary Hardy, an education expert and consultant. I've worked with students of all ages and backgrounds, and I love helping them unlock their full potential. I'm also a big believer in lifelong learning- there's always something new to learn!
I got my start in education as a teacher, working with students in grades K-12. After several years of teaching, I transitioned into the world of educational consulting. I've since worked with schools and districts all over the country, helping them improve their curriculums and instruction methods.
I'm passionate about helping people achieve their dreams, and I believe that education is the key to unlocking everyone's potential. Thanks for reading!
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Can You Solve the 10 Hardest Logic Puzzles Ever Created?
So you think you are clever, right? Then here is your chance to pit your brain against some of the world's hardest logic puzzles ever created. After having created number puzzles like Calcudoku and Killer Sudoku for many years, I decided to try and find the most challenging ones out there. Every once in a while I added a new type of puzzle, until I ended up with a list of 10.
In the following list you will find both familiar puzzles and games such as Sudoku and Calcudoku as well as lesser known ones such as the Bongard Problem and Fill-a-Pix. Some of these puzzles can be solved right on this page while others can be downloaded or reached elsewhere. All of them, however, are promised to test your solving skills to the absolute limit and keep you busy for hours, if not days.
Find an even harder puzzle? Be sure to let me know! For more information about this project and other logic puzzles visit my website Calcudoku.org
1. The World's Hardest Sudoku
Sudoku is easily the most played and most analyzed puzzle in the world, so coming up with the hardest one is no mean feat. In 2012, Finnish mathematician Arto Inkala claimed to have created the "World's Hardest Sudoku".
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According to the British newspaper The Telegraph , on the difficulty scale by which most Sudoku grids are graded, with one star signifying the simplest and five stars the hardest, the above puzzle would "score an eleven". More information on how Inkala's puzzles are rated is on his website .
2. The Hardest Logic Puzzle Ever
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
American philosopher and logician George Boolos invented the above riddle, published in the Harvard Review of Philosophy in 1996, and called it "The Hardest Logic Puzzle Ever". The original article can be downloaded here . You can read about making this puzzle even harder on the Physics arXiv Blog .
3. The World's Hardest Killer Sudoku
A Killer Sudoku is very similar to a Sudoku, except that the clues are given as groups of cells + the sum of the numbers in those cells. From a large number of highest rated puzzles at Calcudoku.org, I measured what percentage of puzzlers solved them on the day they were published. Easily the hardest was the Killer Sudoku shown above, published on the 9th of November 2012. You can solve this puzzle right here .
4. The Hardest Bongard Problem
This type of puzzle first appeared in a book by Russian computer scientist Mikhail Moiseevich Bongard in 1967. They became more widely known after Douglas Hofstadter , an American professor of cognitive science, mentioned them in his book "Gödel, Escher, Bach". To solve the above puzzle, published on Harry Foundalis' website, you have to find a rule that the 6 patterns on the left hand side conform to. The 6 patterns on the right do not conform to this rule. For example, the first problem on this page has as a solution: all patterns on the left are triangles.
5. The Hardest Calcudoku Puzzle
A Calcudoku is similar to a Killer Sudoku, except that (1) any operation can be used to compute the result of a "cage" (not only addition), (2) the puzzle can be any square size, and (3) the Sudoku rule of requiring the numbers 1..9 in each 3×3 set of cells does not apply. Calcudoku was invented by Japanese math teacher Tetsuya Miyamoto , who called it "Kashikoku naru" ("smartness").
Identified in the same way as the Killer Sudoku presented in this article, the hardest Calcudoku was a 9×9 puzzle published on April 2, 2013, which only 9.6% of the regular puzzlers at Calcudoku.org managed to solve. You can give it a try right here . If you're not up for solving it yourself, check out this step-by-step solving analysis by "clm".
6. The Hardest "Ponder this" Puzzle
Design a storage system that encodes 24 information bits on 8 disks of 4 bits each, such that: 1. Combining the 8*4 bits into a 32 bits number (taking a nibble from each disk), a function f from 24 bits to 32 can be computed using only 5 operations, each of which is out of the set {+, -, *, /, %, &, |, ~} (addition; subtraction, multiplication; integer division, modulo; bitwise-and; bitwise-or; and bitwise-not) on variable length integers. In other words, if every operation takes a nanosecond, the function can be computed in 5 nanoseconds. 2. One can recover the original 24 bits even after any 2 of the 8 disks crash (making them unreadable and hence loosing 2 nibbles)
IBM Research has been publishing very challenging monthly puzzles since May 1998 on their Ponder this page. Judging from the number of solvers for each, the hardest number puzzle is the one shown above, published in April 2009. If you need some clues visit this page .
7. The Hardest Kakuro Puzzle
Kakuro puzzles combine elements of Sudoku, logic, crosswords and basic math into one. The object is to fill all empty squares using numbers 1 to 9 so the sum of each horizontal block equals the clue on its left, and the sum of each vertical block equals the clue on its top. In addition, no number may be used in the same block more than once.
Those in the know tell me that the Absolutely Nasty Kakuro Series by Conceptis Puzzles has the world's hardest Kakuro puzzles. Gladly, the guys at Conceptis have produced the above even nastier Kakuro specimen, especially for this article. Play this puzzle online here .
8. Martin Gardner's Hardest Puzzle
A number's persistence is the number of steps required to reduce it to a single digit by multiplying all its digits to obtain a second number, then multiplying all the digits of that number to obtain a third number, and so on until a one-digit number is obtained. For example, 77 has a persistence of four because it requires four steps to reduce it to one digit: 77-49-36-18-8. The smallest number of persistence one is 10, the smallest of persistence two is 25, the smallest of persistence three is 39, and the smaller of persistence four is 77. What is the smallest number of persistence five?
Martin Gardner (1914-2010) was a popular American mathematics and science writer specializing in recreational mathematics, but with interests encompassing micromagic, stage magic, literature, philosophy, scientific skepticism and religion ( Wikipedia ). In his book The Colossal Book of Short Puzzles and Problems puzzles in many categories are listed in order of difficulty. The above is the hardest puzzle from the "Numbers" chapter.
9. The Most Difficult Go Problem Ever
Go is a board game for two players that originated in China more than 2,500 years ago. The game is noted for being rich in strategy despite its relatively simple rules ( Wikipedia ). The above problem is considered to be the hardest ever and is said to have taken 1000 hours to solve by a group of high level students. Solutions and many references can be found on this page .
10. The Hardest Fill-a-Pix Puzzle
Fill-a-Pix is a Minesweeper-like puzzle based on a grid with a pixilated picture hidden inside. Using logic alone, the solver determines which squares are painted and which should remain empty until the hidden picture is completely exposed. Advanced logic Fill-a-Pix such as the one above contain situations where two clues simultaneously affect each other as well as the squares around them making these puzzles extremely hard to solve.
Fill-a-Pix was invented by Trevor Truran , a former high-school math teacher and the editor of Hanjie and several other famed British magazines published by Puzzler Media. For Fill-a-Pix solving rules, advanced solving techniques and more about the history of this puzzle check the Get started section on conceptispuzzles.com. This ultra-hard puzzle was generated by Conceptis especially for this article and can be played online here .
This article originally appeared on Conceptis Puzzles and is reproduced here with kind permission. Conceptis is the leading supplier of logic puzzles to printed and electronic gaming media all over the world. On average, more than 20 million Conceptis puzzles are solved each day in newspapers and magazines, online and on mobile platforms across the world.
Patrick Min is a freelance scientific programmer. He specializes in geometry software, but has worked in many other areas, such as search engine technology, acoustic modelling, and information security. He has published several papers and open/closed-source software across these subjects. Patrick holds a Master's degree in Computer Science from Leiden University, the Netherlands, and a Ph.D. in Computer Science from Princeton University. He is also a puzzle enthusiast, devising math puzzles for his father since the age of 7. This continues to this date, with dad solving his son's Calcudoku puzzles. Patrick lives in London.
Top art by David Masters under Creative Commons license.
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These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It's called a...
For decades, a math puzzle has stumped the smartest mathematicians in the world. x3+y3+z3=k, with k being all the numbers from one to 100, is a Diophantine equation that's sometimes known as...
In 2019, mathematicians finally solved a hard math puzzle that had stumped them for decades. It's called a Diophantine Equation, and it's sometimes known as the "summing of three cubes": Find...
In 1961, the American mathematician Stephen Smale showed that the conjecture is true for n ≥ 5; in 1983, the American mathematician Michael Freedman showed that it is true for n = 4, and in 2002,...
In 2000, the Clay Mathematics Institute, a non-profit dedicated to "increasing and disseminating mathematical knowledge," asked the world to solve seven math problems and offered $1,000,000 to ...
ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by a x − 2 (so you can get rid of the fraction). When you multiply each side by a x − 2, you should have: 24 x 2 + 25 x − 47 = ( − 8 x − 3) ( a x − 2) − 53 You should then multiply ( − 8 x − 3) and ( a x − 2) using FOIL.
With that in mind, we are going to take a look at 6 of the most difficult unsolved math problems in the world. 1.Goldbach Conjecture Let's start our list with an extremely famous and easy-to-understand problem. First, take all the even natural numbers greater than 2 (e.g. 4, 6, 8, 10, 12…).
Mathematicians worldwide hold the Riemann Hypothesis of 1859 (posed by German mathematician Bernhard Riemann (1826-1866)) as the most important outstanding maths problem. The hypothesis states that all nontrivial roots of the Zeta function are of the form (1/2 + b I). All records listed on our website are current and up-to-date.
Mike Dunning/Getty 1. Separatrix Separation A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other...
Equation Four: Solved To determine whether the number 2 (2^127)-1 - 1 is a prime number, we first need to calculate its value. The expression 2 (2^127) can be simplified as follows: 2 (2^127) = 2 * 2^127 = 2^128 Therefore, the expression 2 (2^127)-1 - 1 can be written as 2^128 - 1 - 1. We can then simplify this further to get:
The simplest math problem no one can solve 33,476,338 Views 7,833 Questions Answered Best of Web; Let's Begin… The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. ...
So far, so simple, and it looks like something you would have solved in high school algebra. But here's the problem. Mathematicians haven't ever been able to solve the Beale conjecture, with x, y, and z all being greater than 2. For example, let's use our numbers with the common prime factor of 5 from before…. 5 1 + 10 1 = 15 1. but. 5 2 + 10 ...
Interestingly, these hard math problems can be solved easily with basic math operators. Ready to test your math skills? A bat and a ball cost one dollar and ten cents in total. The bat costs a dollar more than the ball. How much does the ball cost? Answer: The ball costs 5 cents. How did that happen?
Answer (1 of 264): This Is The Hardest Math Problem In The World What is the hardest math problem in the world? The answer to that question is tricky. "Difficulty" is a subjective metric and what is difficult for some may not be difficult for others. Some math problems, such as the infamous ques...
The ACT arranges its questions in order of ascending difficulty. As a general rule of thumb, questions 1-20 will be considered "easy," questions 21-40 will be considered "medium-difficulty," and questions 41-60 will be considered "difficult."
Hardest math problem ever solved. One tool that can be used is Hardest math problem ever solved. order now. These Are the 10 Hardest Math Problems Ever Solved The Riemann hypothesis is one of the Millenium Prize Problems, a list of unsolved math problems compiled by the Clay Institute. The Clay Institute has offered a
When we recently wrote about the toughest math problems that have been solved, we mentioned one of the greatest achievements in 20th century math: the solution to Fermat's Last Theorem. It was solved by Sir Andrew Wiles, using Elliptic Curves. So you could call this a very powerful new branch of math.
Of the original seven Millennium Prize Problems listed by the Clay Mathematics Institute in 2000, six remain unsolved to date: [6] Birch and Swinnerton-Dyer conjecture Hodge conjecture Navier-Stokes existence and smoothness P versus NP Riemann hypothesis Yang-Mills existence and mass gap
In 1988, the Australian Olympiad officials decided to throw a massive curveball to the kids on the final day of competition, and it's gone down in history as one of the toughest problems out there. Just to give you an idea of how tough it was, Australian-American mathematician Terence Tao - recipient of the 2006 Fields Medal (the mathematician ...
These Are the 10 Toughest Math Problems Ever Solved The Collatz Conjecture. Dave Linkletter. Goldbach's Conjecture Creative Commons. The Twin Prime Conjecture. The Riemann Hypothesis. The Birch and Swinnerton-Dyer Conjecture. The Kissing Number Problem. The Unknotting Problem. The Large Cardinal Project. What is the hardest math thing to learn? 1. […]
239,885 views Feb 21, 2019 This question stumped some of the smartest maths students in Australia, and there's no way I would have solved it in an exam! Can you figure it out? Dislike...
Question 2: Answer: F. Category: Number and Quantity. Here's how to solve it: Questions like these are a mere matter of calculating slope. Know that since the sides of a rectangle are parallel, the slope between ( − 1, − 1) and ( 2, 1) will be the same as the one between the fourth vertex and ( 6, − 5).
The above problem is considered to be the hardest ever and is said to have taken 1000 hours to solve by a group of high level students. Solutions and many references can be found on this page . 10.