10 Hardest Math Problem In The World: Solved And Unsolved [With SAT Math Section]

Interested in mathematics? Want to test yourself on the hardest SAT Math questions? Do you want to know what the hardest math problem in the world is? The mysterious world of mathematics is filled with puzzling problems that can stump even the most seasoned mathematicians.

This article will present the world’s 10 hardest math problems, both solved problems and unsolved problems that continue to stump the experts. At the same time, we’ve also compiled the 5 hardest Math problems on SAT for you, with strategies and answer explanations for each question.

What are the Hardest Math Problems?

Mathematical problems such as the Poincaré Conjecture and Fermat’s Last Theorem took centuries to solve. However, other problems, such as the Riemann hypothesis and Goldbach’s conjecture, continue to baffle mathematicians and inspire new generations to find solutions. It is a basic fact in number theory that there are infinitely many primes, which serves as a foundation for various conjectures related to prime numbers. One such well-known conjecture is the Twin Prime Conjecture, which questions whether there are infinitely many prime numbers that differ by 2.

Next, we take a look at some of the 10 hardest math problems. Many mathematical problems have taken mathematicians decades or even centuries to solve, while others remain unsolved.

The table of the 10 hardest math problems is as follows:

Let’s take a look at the ten most difficult math problems through the table.

10 Hardest Math Problems	Status
The Collatz Conjecture	Unsolved
Goldbach’s Conjecture	Unsolved
The twin prime conjecture	Unsolved
The Four Color Theorem	Solved
Riemann Hypothesis	Unsolved
The existence of odd perfect numbers	Unsolved
The Poincaré Conjecture	Solved
The solitary number problem	Unsolved
The Birch and Swinnerton-Dyer Conjecture	Unsolved
Hodge Conjecture	Unsolved

Hardest Math Problem Ever with Answer [Solutions & Examples]

1. The Poincaré Conjecture

In 1904 the French mathematician Henri Poincaré asked if the three-dimensional sphere is characterized as a unique simply connected three manifold. This question, the Poincaré conjecture, was a special case of Thurston’s geometrization conjecture.

• Suppose you have a sphere. You can tie a loop of string anywhere on the string and pull the ends to close the loop. But you can’t do that on a torus (doughnut) if the string goes around through the hole.
• Poincare’s conjecture says that the sphere is the only three-dimensional shape that has this loop-tightening property for every possible loop on its surface. It was proved by Grigori Perelman 20 years ago. In simple terms, he showed that every shape meeting the problem’s criteria can be stretched and shaped into a 3-sphere.

2. The Four Color Theorem

• The four-color theorem, also known as the four-color map theorem: if some adjoining finite regions are delineated in the plane, then these regions can be colored with four colors, such that every two adjoining regions are colored differently; another common way of saying this is that every map of an exclusion-free enclave can be colored with no more than four colors, and no two adjoining regions will have the same color.

The Four Color Conjecture was first stated just over 150 years ago and finally proved in 1976. It is an outstanding example of how old ideas combine with discoveries and techniques in different fields of mathematics to provide new approaches to a problem. It is also an example of how a simple problem was thought to be ‘solved’ but then became more complex, and it is the first spectacular example where a computer was involved in proving a mathematical theorem.

Solution Example: The Four Color Theorem was proven with computer assistance, checking numerous configurations to show that four colors are sufficient. If you want to prove it practically, try coloring a map using only four colors; you’ll find it’s always possible without adjacent regions sharing the same color.

Unsolvable Hardest Math Problem in the World

1. The Collatz Conjecture

The Collatz Conjecture is a simple but unproven mathematical conjecture that proposes a process about sequences of integers. It was proposed by German mathematician Lothar Collatz in 1937 and is described as follows:

• The function f(n) in the figure above, which cuts even numbers in half, cuts odd numbers in triples and then adds to 1, ends up with all of the numbers we examined being 1. The conjecture is that this is true of all natural numbers (positive integers from 1 to infinity).
• The conjecture deals with two simple repeating operations performed on any given positive integer and asks if it will eventually transform the given integer into 1. If the given integer is even, it will be divided by 2. If odd, it will be multiplied by 3, and have 1 added to it. Hence the iconic (3n + 1) name.
• For example, the integer 16. It is even, thus once divided by 2, it becomes 8. 8 being even, will be divided again, becoming 4, then 2, and eventually 1.
• In theory, no matter what positive integer the operations are performed on, it will always transform into 1.

Over the years, many mathematicians have attempted to unravel the mystery of this conjecture, but it has remained an enigma. Many mathematicians have suggested that this problem may even be out of the reach of present-day mathematics. 

Despite the numerous efforts invested in exploring this conjecture, it remains unsolved, and the mathematical community continues to grapple with its intricacies.

2. Goldbach’s Conjecture

The Goldbach conjecture is one of the most captivating mysteries in mathematics. It was proposed by the German mathematician Christian Goldbach in 1742.

• Goldbach’s Conjecture shows that every even natural number greater than 2 can be expressed as the sum of two prime numbers.
• For example: 16 = 3 + 13
• The conjecture was first proposed by Christian Goldbach on June 7, 1742, in a letter to Leonhard Euler. In this letter, Goldbach presented the idea and conjectured that every integer greater than 2 could be expressed as the sum of three prime numbers.
• Euler wrote back that the first part of Goldbach’s conjecture was highly probable, noting that “every even integer is the sum of two prime numbers”, but he was unable to provide proof.

Significant progress has been made in understanding this conjecture over a long period. For example, Nils Pipping verified that n = 100,000 in 1938, and later, with the advent of computers, T. Oliveira e Silva conducted distributed computer searches and by 2013 confirmed the conjecture that n is less than or equal to 4 × 1018 (and repeated the proof for n up to 4 × 1017).

However, a complete and rigorous proof for all even integers greater than 2 remains out of reach.

3. The twin prime conjecture

Proving the twin prime conjecture is a long outstanding problem in number theory. The twin prime conjecture was first formulated by de Polignac in 1849.

• De Polignac argued that for every natural number k, there is an infinite number of primes p such that p+2k is also prime. The case k = 1 is the one we are interested in, the Twin Primes Conjecture.
• It is not surprising that the twin prime conjecture revolves around twin prime numbers. These are prime numbers that are either 2 less or 2 more than another prime number, forming pairs of prime numbers such as (5, 7) or (13, 15).
• The conjecture states that there are an infinite number of prime numbers p such that p + 2 is also prime.

In 2013, Yitang Zhang’s (a Chinese-American mathematician primarily working on number theory) research took an important step towards proving the existence of infinitely many twin prime numbers. His research showed that there exists a finite upper bound – 70 million – for which gaps between pairs of prime numbers exist infinitely often.

By April 2014, this limit (the gap between two prime numbers) had shrunk to 246, indicating significant progress in understanding twin prime numbers.

4. Riemann Hypothesis

The Riemann Hypothesis, formulated by Bernhard Riemann in 1859, is a central problem in number theory that discusses the distribution of prime numbers.

• The hypothesis focuses on the zeros of Riemann’s zeta function. Building on the work of Swiss mathematician Leonhard Euler, Riemann assumed that all non-trivial zeros of this zeta function lie on a critical line in the complex plane, the critical line Re(s) = 0.5.
• The definition of this function is complex in itself, and the infinity of its zeros adds to the challenge. The proof must convincingly show that all zeros cannot deviate from the expected line, which requires a deep understanding of complex analysis and number theory.

The solution of the Riemann hypothesis will be a landmark achievement in mathematics, especially in the field of cryptography, which is crucial for, among other things, Internet security. Confirmation of the hypothesis will also greatly improve our understanding of prime numbers and will validate many mathematical papers that currently take the Riemann hypothesis as a given, thus solidifying a wide range of mathematical theories.

The main challenge in solving the Riemann Hypothesis is that mathematicians simply have not developed the proper tools to tackle the task. Since the problem uses several seemingly disparate parts of mathematics, finding an appropriate restatement of the problem to match the tools currently available to mathematicians seems to be one of the biggest stumbling blocks to solving the Riemann hypothesis.

5. The existence of odd perfect numbers

The existence of odd perfect numbers is a deep unsolved mathematical mystery. In math, a perfect number is a positive integer “n” that equals the sum of all divisors except the number itself. In other words: n = 1 + 2 + 3 + … + (n-1)

A famous example of a perfect number is 28. The divisors of 28 are 1, 2, 4, 7 and 14.

1 + 2 + 4 + 7 + 14 = 28.

However, the existence of odd perfect numbers remains uncertain.

In 1496, Jacques Lefebvre made the point that all perfect numbers can be generated according to Euclid’s law. This implied that there could be no odd perfect numbers, setting the stage for centuries of speculation.

Recently, Carl Pomerance presented a heuristic argument that the existence of odd perfect numbers is highly unlikely. This argument has increased skepticism about their existence.

6. The solitary number problem

The solitary number problem delves into the field of solitary numbers, which are integers that do not have any “friends”; in the mathematical sense (e.g., they do not share a common relationship with any other number). Friendly numbers are numbers that have the same abundance index (the ratio of the sum of the number’s divisors to the number itself).

• Solitary numbers include prime numbers, prime powers, and numbers for which the greatest common divisor of the number and the sum of its divisors (expressed as sigma(n)) equals 1.
• For example, the number 5 is a solitary number. The divisors of 5 are 1 and 5, and their sum is 6. The greatest common divisor of 5 and 6 is 1.

While it is possible to prove the solitariness of some numbers by examining their properties, proving the solitariness of others is challenging. For example, numbers like 10, 15, and 20 are believed to be solitary numbers, but providing conclusive proof has remained elusive.

The concept of solitary numbers has fascinated mathematicians for many years. While prime numbers are well-known solitary numbers, other integers also exhibit solitary properties, even if their greatest common divisor with sigma(n) is not 1.

In 2022, Sourav Mandal shed light on the potential nature of 10’s friend, proposing a specific form it must follow if it exists, adding an intriguing layer to the problem. Furthermore, examples like 24, classified as friendly, and possessing 91,963,648 as its smallest friend, illustrate the diversity in the classification of numbers as friendly or solitary.

7. The Birch and Swinnerton-Dyer Conjecture

• The Birch and Swinnerton-Dyer conjecture is a far-reaching and complex problem in number theory, focusing on elliptic curves. The conjecture relates the number of rational solutions (points whose two coordinates are rational numbers) on an elliptic curve to certain features of the L-function associated with that curve.
• Essentially, it predicts that the behavior of this L-function at certain values ​​can indicate whether the curve has a finite or infinite number of rational points. The conjecture is notoriously tricky to solve due to its deep connections to various complex mathematical concepts such as L-functions, elliptic curves, and modular forms.

While significant progress has been made in understanding specific cases and aspects of the conjecture, a general proof or disproof remains elusive. A solution to the conjecture would have a significant impact on several areas of mathematics, particularly in number theory and algebraic geometry. It would increase our understanding of elliptic curves, which are central to many areas of mathematics, including cryptography, and could lead to advances in digital security and new cryptographic techniques.

8. Hodge Conjecture

The Hodge conjecture was proposed by William Hodge in 1941 and involves algebraic geometry. It shows a fundamental relationship between collections of simple geometric pieces, called algebraic cycles, and complex shapes of certain “nice” spaces, called projective algebraic varieties. The Hodge conjecture asserts that these algebraic cycles can approximate the shapes of these varieties.

In short, solving this problem is like putting together a very complex, abstract puzzle. The Hodge conjecture is very difficult because of its complex nature and deep connections to various areas of mathematics. It requires a deep understanding of algebraic geometry, complex geometry, and topology.

A solution to the Hodge conjecture could advance our understanding of higher-dimensional mathematical structures and could have applications in areas such as string theory and other parts of theoretical physics.

5 Hardest Math Problems on SAT with Answers

In addition to the great math problems that have been solved and unsolved above, we’ve prepared some things you might want to know about the SAT math section. Want to test the hardest SAT math problems? Want to know what makes them so difficult and how to best solve them?

These are all SAT math problems from the College Board’s SAT practice tests, which means that understanding them is one of the best ways to study for those who want to prepare for the SAT or just aim for perfection.

Question 1.

The equation above shows the relationship between temperature F, measured in degrees Fahrenheit, and temperature C, measured in degrees Celsius. Based on this equation, which of the following must be true?

1. A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of 5/9 degree Celsius.
2. A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.
3. A temperature increase of 5/9 degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

A) 1 only
B) 2 only
C) 3 only
D) 1 and 2 only

ANSWER EXPLANATION: Think of the equation as an equation for a line y = mx + b, where in this case C = 5/9(F – 32)

You can see the slope of the graph is 5/9, which means that for an increase of 1 degree Fahrenheit, the increase is 5/9 of 1 degree Celsius.

C = 5/9(F) , C = 5/9(1) = 5/9

Therefore, statement I is true. This is the equivalent to saying that an increase of 1 degree Celsius is equal to an increase of 9/5 degrees Fahrenheit.

C = 5/9(F) , 1 = 5/9(F) , (F) = 9/5

Since 9/5 = 1.8, statement 2 is true. The only answer that has both statement 1 and statement 2 as true is D, but if you have time and want to be absolutely thorough, you can also check to see if statement 3 (an increase of 5/9 degree Fahrenheit is equal to a temperature increase of 1 degree Celsius) is true: C = 5/9(F) , C = 5/9(5/9) , C = 25/81(which is ≠ 1)

An increase of 5/9 degree Fahrenheit leads to an increase of 25/81, not 1 degree, Celsius, and so Statement 3 is not true.

The final answer is D.

Question 2.

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax − 2 (so you can get rid of the fraction). When you multiply each side by ax − 2, you should have: 24x²+ 25x – 47 = (-8x – 3) (ax – 2) – 53

You should then multiply (-8x – 3) and (ax – 2) using FOIL.

24x²+ 25x – 47 = -8ax² – 3ax + 16x + 6 – 53

Then, reduce on the right side of the equation 24x²+ 25x – 47 = -8ax² – 3ax + 16x – 47

Since the coefficients of the x² -term have to be equal on both sides of the equation, -8a = 24, or a = -3. The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.

The final answer is B.

Question 3.

The incomplete table above summarizes the number of left-handed students and right-handed students by gender for the eighth grade students at Keisel Middle School. There are 5 times as many right-handed female students as there are left-handed female students, and there are 9 times as many right-handed male students as there are left-handed male students.

If there is a total of 18 left-handed students and 122 right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.)

A) 0.410

B) 0.357

C) 0.333

D) 0.250

ANSWER EXPLANATION: In order to solve this problem, you should create two equations using two variables (x and y) and the information you’re given. Let x be the number of left-handed female students and let y be the number of left-handed male students. Using the information given in the problem, the number of right-handed female students will be 5x and the number of right-handed male students will be 9y. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

x + y = 18

5x + 9y = 122

When you solve this system of equations, you get x = 10 and y = 8. Thus, 5*10, or 50, of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is 50/122, which to the nearest thousandth is 0.410.

The final answer is A.

Question 4.

Little’s law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

ANSWER EXPLANATION: Since the question states that Little’s law can be applied to any single part of the store (for example, just the checkout line), then the average number of shoppers, N, in the checkout line at any time is N = rT, where r is the number of shoppers entering the checkout line per minute and T is the average number of minutes each shopper spends in the checkout line. Since 84 shoppers per hour make a purchase, 84 shoppers per hour enter the checkout line. However, this needs to be converted to the number of shoppers per minute (in order to be used with T=5). Since there are 60 minutes in one hour, the rate is 84

shoppers per hour/60 minutes =1.4 shoppers per minute. Using the given formula with r=1.4 and T=5 yields

N=rt=(1.4)(5)=7

Therefore, the average number of shoppers, N, in the checkout line at any time during business hours is 7.

The final answer is 7.

Question 5.

The owner of the Good Deals Store opens a new store across town. For the new store, the owner estimates that, during business hours, an average of 90 shoppers per hour enter the store and each of them stays an average of 12 minutes. The average number of shoppers in the new store at any time is what percent less than the average number of shoppers in the original store at any time? (Note: Ignore the percent symbol when entering your answer. For example, if the answer is 42.1%, enter 42.1)

ANSWER EXPLANATION: According to the original information given, the estimated average number of shoppers in the original store at any time (N) is 45. In the question, it states that, in the new store, the manager estimates that an average of 90 shoppers per hour (60 minutes) enter the store, which is equivalent to 1.5 shoppers per minute (r). The manager also estimates that each shopper stays in the store for an average of 12 minutes (T). Thus, by Little’s law, there are, on average, N=rT=(1.5)(12)=18 shoppers in the new store at any time. This is

45−18/45*100=60

percent less than the average number of shoppers in the original store at any time.

The final answer is 60.

Common Themes in Hardest Math Problems

Math problems often involve complex mathematical concepts, abstract thinking, and critical reasoning. They require a deep understanding of mathematical principles, theories, and formulas.

Here are some typical examples of math problems:

1. Millennium Prize Problems: These problems are seven math problems identified by the Clay Mathematics Institute as the most important unsolved problems in mathematics. They include problems such as the Riemann Hypothesis, the P vs. NP problem, and the Navier-Stokes equations. Solving any one of these problems would represent a major breakthrough in mathematics.
2. SAT Math Section: The SAT Math section is a challenging test of math knowledge and problem-solving skills. It includes problems that require a deep understanding of math concepts, such as algebra, geometry, and trigonometry. Mastering this section requires knowledge and the ability to apply mathematical principles to solve complex problems.
3. The Difficult SAT Math Questions: These questions are a set of challenging math problems designed to test a student’s mathematical knowledge and problem-solving skills. They include problems such as the ones listed in the article “The Hardest SAT Math Questions Ever”. These questions often require creative thinking and a deep understanding of mathematical concepts to find the correct answer.

By understanding these common themes, students, teachers, and anyone interested in math can be better prepared to solve difficult math problems, whether preparing for the SAT or working on some of the most challenging problems in math.

FAQs on the Hardest Math Problems

1. What is the most difficult math section on the SAT?

The most challenging math section of the SAT is often considered the “Problem Solving and Data Analysis” section.

According to the College Board, many students typically score lower on this section than on other sections. This section requires strong skills in interpreting data, understanding ratios, and applying math concepts to real-world scenarios. Recent SAT score data shows that many students struggle with this type of question, affecting their overall score.

2. What are 7th grade math problems? [with examples]

7th grade math problems often involve concepts such as solving equations, calculating percentages, or understanding geometry. This level of math builds on prior knowledge and prepares students for more advanced topics such as algebra and geometry in high school. 7th grade math problems build on students’ reasoning skills and understanding of key mathematical concepts.

Here are some examples of 7th grade math problems:

1. Isabella got 16 out of 40 questions wrong on her quiz. What percent did she get correct?Solution: 16/40 can be simplified to 2/5​, which is equivalent to 40/100​ or 40%. If Isabella got 40% incorrect, she got 60% correct (100-40=60).
2. Three out of every five students are wearing jeans. If there are 20 students in all, how many are not wearing jeans?Solution: From the last problem, we saw that 3/5​ is the same as the 12 students wearing jeans. If there are 20 students total, we can subtract the 12 wearing jeans from the 20 total to find that 8 are not wearing jeans. We could also set up this proportion and solve to get 8.2/5 = x/20
3. Madison measured this angle with her protractor and said “It is 60°.” Without measuring the angle, Bella said she could tell Madison’s answer was incorrect. How did Bella know this?

Solution: Bella knew this angle could not be 60° because this angle is obtuse but a 60° angle is acute.

3. What are the prime numbers and their significance?

To help you better understand the above math problems, we have given some relevant knowledge tips.

First, prime numbers play a crucial role in mathematics, particularly in number theory. A prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself. The study of prime numbers is essential in understanding many mathematical concepts, including the distribution of prime numbers, the prime number theorem, and the twin prime conjecture.

Moreover, prime numbers are the building blocks of the integers, much like atoms are the building blocks of matter. They are fundamental in various branches of mathematics and have intrigued mathematicians for centuries. The prime number theorem, for instance, describes the asymptotic distribution of prime numbers among the positive integers, providing a deep insight into their occurrence.

Finally, the twin prime conjecture, another fascinating problem, posits that there are infinitely many pairs of prime numbers (twin primes) that differ by exactly two. Despite significant progress, this conjecture remains unproven, continuing to challenge and inspire mathematicians.

Conclusion

These are the top 10 hardest math problems in the world and the five hardest SAT math problems. Some of them have been solved perfectly, while the complexity of some of them still poses a challenge to the academic world. For math enthusiasts, this is an arena where they can continue to hone their problem-solving skills.

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