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Want to test yourself against the most difficult SAT math questions? Want to know what makes these questions so difficult and how best to solve them? If you're ready to really sink your teeth into the SAT math section and have your sights set on that perfect score, then this is the guide for you.

We've put together what we believe to be the 15 most difficult questions for the current SAT , with strategies and answer explanations for each. These are all hard SAT Math questions from College Board SAT practice tests, which means understanding them is one of the best ways to study for those of you aiming for perfection.

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Brief Overview of SAT Math

The third and fourth sections of the SAT will always be math sections . The first math subsection (labeled "3") does not allow you to use a calculator, while the second math subsection (labeled as "4") does allow the use of a calculator. Don't worry too much about the no-calculator section, though: if you're not allowed to use a calculator on a question, it means you don't need a calculator to answer it.

Each math subsection is arranged in order of ascending difficulty (where the longer it takes to solve a problem and the fewer people who answer it correctly, the more difficult it is). On each subsection, question 1 will be "easy" and question 15 will be considered "difficult." However, the ascending difficulty resets from easy to hard on the grid-ins.

Hence, multiple choice questions are arranged in increasing difficulty (questions 1 and 2 will be the easiest, questions 14 and 15 will be the hardest), but the difficulty level resets for the grid-in section (meaning questions 16 and 17 will again be "easy" and questions 19 and 20 will be very difficult).

With very few exceptions, then, the most difficult SAT math problems will be clustered at the end of the multiple choice segments or the second half of the grid-in questions. In addition to their placement on the test, though, these questions also share a few other commonalities. In a minute, we'll look at example questions and how to solve them, then analyze them to figure out what these types of questions have in common.

But First: Should You Be Focusing on the Hardest Math Questions Right Now?

If you're just getting started in your study prep (or if you've simply skipped this first, crucial step), definitely stop and take a full practice test to gauge your current scoring level. Check out our guide to all the free SAT practice tests available online and then sit down to take a test all at once.

The absolute best way to assess your current level is to simply take the SAT practice test as if it were real , keeping strict timing and working straight through with only the allowed breaks (we know—probably not your favorite way to spend a Saturday). Once you've got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate SAT Math score.

If you're currently scoring in the 200-400 or the 400-600 range on SAT Math, your best bet is first to check out our guide to improving your math score to be consistently at or over a 600 before you start in trying to tackle the most difficult math problems on the test.

If, however, you're already scoring above a 600 on the Math section and want to test your mettle for the real SAT, 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 SAT math questions look like and how to solve them. And luckily, that's exactly what we'll do.

WARNING: Since there are a limited number of official SAT practice tests , you may want to wait to read this article until you've attempted all or most of the first four official practice tests (since most of the questions below were taken from those tests). If you're worried about spoiling those tests, stop reading this guide now; come back and read it when you've completed them.

Now let's get to our list of questions (whoo)!

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The 15 Hardest SAT Math Questions

Now that you're sure you should be attempting these questions, let's dive right in! We've curated 15 of the most difficult SAT Math questions for you to try below, along with walkthroughs of how to get the answer (if you're stumped).

No Calculator SAT Math Questions

$$C=5/9(F-32)$$

The equation above shows how temperature $F$, measured in degrees Fahrenheit, relates to a temperature $C$, measured in degrees Celsius. Based on the equation, which of the following must be true?

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

A) I only B) II only C) III only D) I and II only

ANSWER EXPLANATION: Think of the equation as an equation for a line

where in this case

$$C= {5}/{9} (F−32)$$

$$C={5}/{9}F −{5}/{9}(32)$$

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.

$$1= {5}/{9} (F)$$

$$(F)={9}/{5}$$

Since ${9}/{5}$ = 1.8, statement II is true.

The only answer that has both statement I and statement II as true is D , but if you have time and want to be absolutely thorough, you can also check to see if statement III (an increase of ${5}/{9}$ degree Fahrenheit is equal to a temperature increase of 1 degree Celsius) is true:

$$C= {5}/{9} ({5}/{9})$$

$$C= {25} /{81} (\which \is ≠ 1)$$

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

The final answer is D.

The equation ${24x^2 + 25x -47}/{ax-2} = -8x-3-{53/{ax-2}}$ is true for all values of $x≠2/a$, where $a$ is a constant.

What is the value of $a$?

A) -16 B) -3 C) 3 D) 16

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:

$$24x^2 + 25x - 47 = (-8x-3)(ax-2) - 53$$

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

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x + 6 - 53$$

Then, reduce on the right side of the equation

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x - 47$$

Since the coefficients of the $x^2$-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.

If $3x-y = 12$, what is the value of ${8^x}/{2^y}$?

A) $2^{12}$ B) $4^4$ C) $8^2$ D) The value cannot be determined from the information given.

ANSWER EXPLANATION: One approach is to express

$${8^x}/{2^y}$$

so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting $2^3$ for 8 in the numerator of ${8^x}/{2^y}$ gives

$${(2^3)^x}/{2^y}$$

which can be rewritten

$${2^3x}/{2^y}$$

Since the numerator and denominator of have a common base, this expression can be rewritten as $2^(3x−y)$. In the question, it states that $3x − y = 12$, so one can substitute 12 for the exponent, $3x − y$, which means that

$${8^x}/{2^y}= 2^12$$

The final answer is A.

Points A and B lie on a circle with radius 1, and arc ${AB}↖⌢$ has a length of $π/3$. What fraction of the circumference of the circle is the length of arc ${AB}↖⌢$?

ANSWER EXPLANATION: To figure out the answer to this question, you'll first need to know the formula for finding the circumference of a circle.

The circumference, $C$, of a circle is $C = 2πr$, where $r$ is the radius of the circle. For the given circle with a radius of 1, the circumference is $C = 2(π)(1)$, or $C = 2π$.

To find what fraction of the circumference the length of ${AB}↖⌢$ is, divide the length of the arc by the circumference, which gives $π/3 ÷ 2π$. This division can be represented by $π/3 * {1/2}π = 1/6$.

The fraction $1/6$ can also be rewritten as $0.166$ or $0.167$.

The final answer is $1/6$, $0.166$, or $0.167$.

$${8-i}/{3-2i}$$

If the expression above is rewritten in the form $a+bi$, where $a$ and $b$ are real numbers, what is the value of $a$? (Note: $i=√{-1}$)

ANSWER EXPLANATION: To rewrite ${8-i}/{3-2i}$ in the standard form $a + bi$, you need to multiply the numerator and denominator of ${8-i}/{3-2i}$ by the conjugate, $3 + 2i$. This equals

$$({8-i}/{3-2i})({3+2i}/{3+2i})={24+16i-3+(-i)(2i)}/{(3^2)-(2i)^2}$$

Since $i^2=-1$, this last fraction can be reduced simplified to

$$ {24+16i-3i+2}/{9-(-4)}={26+13i}/{13}$$

which simplifies further to $2 + i$. Therefore, when ${8-i}/{3-2i}$ is rewritten in the standard form a + bi, the value of a is 2.

In triangle $ABC$, the measure of $∠B$ is 90°, $BC=16$, and $AC$=20. Triangle $DEF$ is similar to triangle $ABC$, where vertices $D$, $E$, and $F$ correspond to vertices $A$, $B$, and $C$, respectively, and each side of triangle $DEF$ is $1/3$ the length of the corresponding side of triangle $ABC$. What is the value of $sinF$?

ANSWER EXPLANATION: Triangle ABC is a right triangle with its right angle at B. Therefore, $\ov {AC}$ is the hypotenuse of right triangle ABC, and $\ov {AB}$ and $\ov {BC}$ are the legs of right triangle ABC. According to the Pythagorean theorem,

$$AB =√{20^2-16^2}=√{400-256}=√{144}=12$$

Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of $\angle ∠ {F}$ equals the measure of $\angle ∠ {C}$. Therefore, $sin F = sin C$. From the side lengths of triangle ABC,

$$sinF ={\opposite \side}/{\hypotenuse}={AB}/{AC}={12}/{20}={3}/{5}$$

Therefore, $sinF ={3}/{5}$.

The final answer is ${3}/{5}$ or 0.6.

Calculator-Allowed SAT Math Questions

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.

Questions 8 & 9

Use the following information for both question 7 and question 8.

If shoppers enter a store at an average rate of $r$ shoppers per minute and each stays in the store for average time of $T$ minutes, the average number of shoppers in the store, $N$, at any one time is given by the formula $N=rT$. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

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.

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.

Question 10

In the $xy$-plane, the point $(p,r)$ lies on the line with equation $y=x+b$, where $b$ is a constant. The point with coordinates $(2p, 5r)$ lies on the line with equation $y=2x+b$. If $p≠0$, what is the value of $r/p$?

ANSWER EXPLANATION: Since the point $(p,r)$ lies on the line with equation $y=x+b$, the point must satisfy the equation. Substituting $p$ for $x$ and $r$ for $y$ in the equation $y=x+b$ gives $r=p+b$, or $\bi b$ = $\bi r-\bi p$.

Similarly, since the point $(2p,5r)$ lies on the line with the equation $y=2x+b$, the point must satisfy the equation. Substituting $2p$ for $x$ and $5r$ for $y$ in the equation $y=2x+b$ gives:

$5r=2(2p)+b$

$\bi b$ = $\bo 5 \bi r-\bo 4\bi p$.

Next, we can set the two equations equal to $b$ equal to each other and simplify:

$b=r-p=5r-4p$

Finally, to find $r/p$, we need to divide both sides of the equation by $p$ and by $4$:

The correct answer is B , $3/4$.

If you picked choices A and D, you may have incorrectly formed your answer out of the coefficients in the point $(2p, 5r)$. If you picked Choice C, you may have confused $r$ and $p$.

Note that while this is in the calculator section of the SAT, you absolutely do not need your calculator to solve it!

Question 11

A) 261.8 B) 785.4 C) 916.3 D) 1047.2

ANSWER EXPLANATION: The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the SAT Math section:

Volume of a Cone

$$V={1}/{3}πr^2h$$

Volume of a Cylinder

$$V=πr^2h$$

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

$$V_{silo}=π(5^2)(10)+(2)({1}/{3})π(5^2)(5)=({4}/{3})(250)π$$

which is approximately equal to 1,047.2 cubic feet.

Question 12

If $x$ is the average (arithmetic mean) of $m$ and $9$, $y$ is the average of $2m$ and $15$, and $z$ is the average of $3m$ and $18$, what is the average of $x$, $y$, and $z$ in terms of $m$?

A) $m+6$ B) $m+7$ C) $2m+14$ D) $3m + 21$

ANSWER EXPLANATION: Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2, the equations $x={m+9}/{2}$, $y={2m+15}/{2}$, $z={3m+18}/{2}$are true. The average of $x$, $y$, and $z$ is given by ${x + y + z}/{3}$. Substituting the expressions in m for each variable ($x$, $y$, $z$) gives

$$[{m+9}/{2}+{2m+15}/{2}+{3m+18}/{2}]/3$$

This fraction can be simplified to $m + 7$.

Question 13

The function $f(x)=x^3-x^2-x-{11/4}$ is graphed in the $xy$-plane above. If $k$ is a constant such that the equation $f(x)=k$ has three real solutions, which of the following could be the value of $k$?

ANSWER EXPLANATION: The equation $f(x) = k$ gives the solutions to the system of equations

$$y = f(x) = x^3-x^2-x-{11}/{4}$$

A real solution of a system of two equations corresponds to a point of intersection of the graphs of the two equations in the $xy$-plane.

The graph of $y = k$ is a horizontal line that contains the point $(0, k)$ and intersects the graph of the cubic equation three times (since it has three real solutions). Given the graph, the only horizontal line that would intersect the cubic equation three times is the line with the equation $y = −3$, or $f(x) = −3$. Therefore, $k$ is $-3$.

Question 14

$$q={1/2}nv^2$$

The dynamic pressure $q$ generated by a fluid moving with velocity $v$ can be found using the formula above, where $n$ is the constant density of the fluid. An aeronautical engineer users the formula to find the dynamic pressure of a fluid moving with velocity $v$ and the same fluid moving with velocity 1.5$v$. What is the ratio of the dynamic pressure of the faster fluid to the dynamic pressure of the slower fluid?

ANSWER EXPLANATION: To solve this problem, you need to set up to equations with variables. Let $q_1$ be the dynamic pressure of the slower fluid moving with velocity $v_1$, and let $q_2$ be the dynamic pressure of the faster fluid moving with velocity $v_2$. Then

$$v_2 =1.5v_1$$

Given the equation $q = {1}/{2}nv^2$, substituting the dynamic pressure and velocity of the faster fluid gives $q_2 = {1}/{2}n(v_2)^2$. Since $v_2 =1.5v_1$, the expression $1.5v_1$ can be substituted for $v_2$ in this equation, giving $q_2 = {1}/{2}n(1.5v_1)^2$. By squaring $1.5$, you can rewrite the previous equation as

$$q_2 = (2.25)({1}/{2})n(v_1)^2 = (2.25)q_1$$

Therefore, the ratio of the dynamic pressure of the faster fluid is

$${q2}/{q1} = {2.25 q_1}/{q_1}= 2.25$$

The final answer is 2.25 or 9/4.

Question 15

For a polynomial $p(x)$, the value of $p(3)$ is $-2$. Which of the following must be true about $p(x)$?

A) $x-5$ is a factor of $p(x)$. B) $x-2$ is a factor of $p(x)$. C) $x+2$ is a factor of $p(x)$. D) The remainder when $p(x)$ is divided by $x-3$ is $-2$.

ANSWER EXPLANATION: If the polynomial $p(x)$ is divided by a polynomial of the form $x+k$ (which accounts for all of the possible answer choices in this question), the result can be written as

$${p(x)}/{x+k}=q(x)+{r}/{x+k}$$

where $q(x)$ is a polynomial and $r$ is the remainder. Since $x + k$ is a degree-1 polynomial (meaning it only includes $x^1$ and no higher exponents), the remainder is a real number.

Therefore, $p(x)$ can be rewritten as $p(x) = (x + k)q(x) + r$, where $r$ is a real number.

The question states that $p(3) = -2$, so it must be true that

$$-2 = p(3) = (3 + k)q(3) + r$$

Now we can plug in all the possible answers. If the answer is A, B, or C, $r$ will be $0$, while if the answer is D, $r$ will be $-2$.

A. $-2 = p(3) = (3 + (-5))q(3) + 0$ $-2=(3-5)q(3)$ $-2=(-2)q(3)$

This could be true, but only if $q(3)=1$

B. $-2 = p(3) = (3 + (-2))q(3) + 0$ $-2 = (3-2)q(3)$ $-2 = (-1)q(3)$

This could be true, but only if $q(3)=2$

C. $-2 = p(3) = (3 + 2)q(3) + 0$ $-2 = (5)q(3)$

This could be true, but only if $q(3)={-2}/{5}$

D. $-2 = p(3) = (3 + (-3))q(3) + (-2)$ $-2 = (3 - 3)q(3) + (-2)$ $-2 = (0)q(3) + (-2)$

This will always be true no matter what $q(3)$ is.

Of the answer choices, the only one that must be true about $p(x)$ is D, that the remainder when $p(x)$ is divided by $x-3$ is -2.

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You deserve all the naps after running through those questions.

What Do the Hardest SAT Math Questions Have in Common?

It's important to understand what makes these hard questions "hard." By doing so, you'll 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 SAT math errors.

In this section, we'll look at what these questions have in common and give examples of each type. Some of the reasons why the hardest math questions are the hardest math questions is because they:

#1: Test Several Mathematical Concepts at Once

Here, we must deal with imaginary numbers and fractions all at once.

Secret to success: Think of what applicable math you could use to solve the problem, do one step at a time, and try each technique until you find one that works!

#2: Involve a Lot of Steps

Remember: the more steps you need to take, the easier to mess up somewhere along the line!

We must solve this problem in steps (doing several averages) to unlock the rest of the answers in a domino effect. This can get confusing, especially if you're stressed or running out of time.

Secret to success: Take it slow, take it step by step, and double-check your work so you don't make mistakes!

#3: Test Concepts That You Have Limited Familiarity With

For example, many students are less familiar with functions than they are with fractions and percentages, so most function questions are considered "high difficulty" problems.

If you don't know your way around functions , this would be a tricky problem.

Secret to success: Review math concepts that you don't have as much familiarity with such as functions . We suggest using our great free SAT Math review guides .

#4: Are Worded in Unusual or Convoluted Ways

It can be difficult to figure out exactly what some questions are asking , much less figure out how to solve them. This is especially true when the question is located at the end of the section, and you are running out of time.

Because this question provides so much information without a diagram, it can be difficult to puzzle through in the limited time allowed.

Secret to success: Take your time, analyze what is being asked of you, and draw a diagram if it's helpful to you.

#5: Use Many Different Variables

With so many different variables in play, it is quite easy to get confused.

Secret to success: Take your time, analyze what is being asked of you, and consider if plugging in numbers is a good strategy to solve the problem (it wouldn't be for the question above, but would be for many other SAT variable questions).

The Take-Aways

The SAT is a marathon and the better prepared you are for it, the better you'll feel on test day. Knowing how to handle the hardest questions the test can throw at you will make taking the real SAT 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 problems. As you continue to study, always adhere to the proper timing guidelines 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 SAT . 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 in the SAT math section and then note which sections were particular difficulty for you. Next, take a gander at our individual math guides to help you shore up any of those weak areas.

Running out of time on the SAT math section? Our guide will help you beat the clock and maximize your score .

Aiming for a perfect score? Check out our guide on how to get a perfect 800 on the SAT math section , written by a perfect-scorer.

Want to improve your SAT score by 160 points?

Check out our best-in-class online SAT prep classes . We guarantee your money back if you don't improve your SAT score by 160 points or more.

Our classes are entirely online, and they're taught by SAT experts . If you liked this article, you'll love our classes. Along with expert-led classes, you'll get personalized homework with thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step, custom program to follow so you'll never be confused about what to study next.

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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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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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These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance

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.

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.

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.

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.

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.

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.

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.

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.

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.

These ten brutally difficult math problems once seemed impossible until mathematicians eventually solved them—even if it took them years, decades, or centuries.

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9 Hardest Calculus Problems Ever You’ll Ever Encounter

Lately, I was teaching one of the brightest students; she asked me what the hardest calculus problem ever was. Her question led me to do deeper research to find.

Mathematics is a constantly evolving field, and new equations and calculations are constantly being discovered. But some problems have posed a challenge for mathematicians for centuries. Here are the

In this blog, I wanted to share what I found; maybe you try solving it. Read on to find some of the most challenging calculus problems and discuss why they are so difficult.

The good news is two of the hardest calculus problems are still unsolved, and there is a reward of $1 million dollars for whoever finds the answers to each problem.

You might also enjoy reading: What Jobs Can You Get With a Mathematics Degree: 9 Best Options .

Table of Contents

1- The Three-Body Problem

The Three-Body Problem is one of the oldest and most famous unsolved problems in mathematics. It was first proposed by Isaac Newton in 1687 and remains unsolved to this day (Source: Scientific American )

The three-body problem deals with understanding the motion of three objects interacting with each other, such as moons orbiting planets or stars in galaxies, given their initial positions and velocities.

It has been particularly difficult for mathematicians due to its chaotic behavior, meaning that small changes in the initial conditions can lead to drastically different outcomes. Additionally, its nonlinearity makes it resistant to traditional mathematical techniques.

When Isaac Newton published his  Principia  in 1687, he asked: “How will two masses move in space if the only force on them is their mutual gravitational attraction? ” Newton formulated the question as a problem solving a system of differential equations . 

Despite these challenges, many researchers have made significant progress on the Three-Body Problem over the years, but it still remains unsolved (Source: Popular Mechanics )

Watch the video below to learn more about the Three-Body Problem.

2- Goldbach’s Conjecture

Christian Goldbach first proposed this conjecture in 1742 and stated that every even number greater than two could be written as the sum of two prime numbers (a prime number is an integer greater than one with no divisors other than itself).

For example, 8 = 3 + 5 or 10 = 7 + 3. While this conjecture seems simple enough at first glance, it has proven surprisingly hard to prove or disprove!

Despite intense effort from mathematicians worldwide over 250 years, Goldbach’s Conjecture remains unproven and stands as one of the greatest open problems in mathematics today.

Further progress on Goldbach’s conjecture emerged in 1973 when the Chinese mathematician Chen Jing Run demonstrated that every sufficiently large even number is the sum of a prime and a number with at most two prime factors .

3- Fermat’s Last Theorem

This theorem dates back to 1637 when Pierre de Fermat wrote down his famous equation without providing any proof or explanation for it in his notebook: “it is impossible to separate a cube into two cubes or a fourth power into two fourth powers or generally any power higher than second into two like powers.”

Fermat’s last theorem, also known as Fermat’s great theorem, is the statement that there exist no natural numbers (1, 2, 3,…) x, y, and z such that x^n + y^n = z^n, in which n is a natural number bigger than 2.

 For instance, if  n  = 3, Fermat’s last theorem says that no natural numbers  x ,  y , and  z  exist such that  x^ 3 +  y  ^3 =  z^ 3. In other words, the sum of two cubes is not a cube (Source: Britannica )

The Fermat’s Last Theorem remained unproven until 1995 when Andrew Wiles finally provided proof using elliptic curves after working on it for seven years (Source: National Science Foundation (NSF) )

This theorem stands as one of the greatest achievements in mathematics and still remains one of the most difficult problems ever tackled by mathematicians worldwide.

Feynman wrote an unpublished 2 page manuscript approaching Fermat’s Last Theorem from a probabilistic standpoint and concluded (before Andrew Wiles’ proof!) that “for my money Fermat’s theorem is true”. Here is the reconstruction of his approach: https://t.co/3GrUNXEfuW pic.twitter.com/sDpUD5JWJF — Fermat’s Library (@fermatslibrary) November 5, 2018

4- The Riemann Hypothesis

The Riemann Hypothesis is perhaps one of the most famous unsolved problems in mathematics today . It states that all non-trivial zeros of the Riemann zeta function have real parts equal to 1/2.

While it has not yet been proven (or disproven), mathematicians have made considerable progress towards solving it using techniques from complex analysis and number theory.

Unfortunately, many mathematicians believe it may never be solved without major mathematics and computer science breakthroughs due to its complexity and difficulty.

If you are looking for ways to make a million dollars by solving math, try solving the Riemann Hypothesis. It is among the  Seven Millennium Prize Problems , with a $1 million reward if you find its solutions.  

If you solve the Riemann Hypothesis tomorrow, it will open an avalanche of further progress. It would be massive news throughout the topics of Number Theory and Analysis. 

I suggest you watch the video below to learn more about the Riemann Hypothesis.

5- The Collatz Conjecture

The Collatz Conjecture is another unsolved mathematical problem that has remained a mystery since its inception in 1937. Intuitively described, it deals with the sequence created by taking any number and, if it is even, dividing it by two, and if it is odd, multiplying by three and adding one .

Every cycle of this algorithm eventually converges to the same number: 1. So far, no one has been able to determine why this happens or why the Collatz Conjecture holds true for all natural numbers (positive integers from 1 to infinity).

This elusive problem has stumped mathematicians for decades and continues to draw researchers to try and solve this head-scratching conundrum.

Despite numerous attempts made to unravel its secrets, the Collatz Conjecture remains as enigmatic as ever, begging us to discover its mystery and open up new doors in the realm of mathematics.

If proved true, the Collatz Conjecture could provide major new insights into our understanding of mathematics and computing algorithms, leading to numerous potential applications.

Undoubtedly, whoever solves this hypothesis will have made one of the great discoveries in mathematics.

The video below discusses the Collatz Conjecture.

6- The Twin Prime Conjecture

In number theory, the Twin Prime conjecture , also known as Polignac’s conjecture, asserts that infinitely many twin primes, or pairs of primes, differ by 2.  As an illustration, 3 and 5, 5 and 7, 11 and 13, and 17 and 19 are considered twin primes. As numbers become larger, primes become less frequent, and twin primes are rarer still.

The Twin Prime Conjecture is an unsolved problem in mathematics that has stumped the best minds for centuries. If the conjecture is true, it will open up a whole new realm of prime numbers, providing new avenues for exploration and even potential applications in cryptography.

However, it has been difficult to prove due to the lack of general patterns for consecutive primes; any pattern made thus far is inconsistent and unreliable at best.

Despite this difficulty, mathematicians remain optimistic about uncovering the answer to this mystery–and when they do, it will surely be a monumental achievement!

I encourage you to watch the video below to learn more about the Twin Prime Conjecture.

7- The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer Conjecture , a crucial unsolved mathematical problem in number theory, remains one of the greatest mysteries of our time. The Birch and Swinnerton-Dyer Conjecture is also among the six unsolved Millennium Prize Problems, meaning that if you solve it, you will be rewarded with one million dollars.

Originally conjectured in the 1960s, this idiosyncratic conjecture has captivated mathematicians ever since. While researchers have gained insight into related topics such as elliptic curves and modular forms, the true complexity of this conjecture has still eluded them.

An elliptic curve is a particular kind of function that can be written in this form y²=x³+ax+b. It turns out that these types of functions have specific properties that explore other math topics, such as Algebra and Number Theory.

As a result, researchers continue to explore new approaches and hope they can one day demonstrate their veracity. Undoubtedly, this intriguing yet tough problem will captivate mathematicians for years to come.

If you are interested in learning more about the Birch-Swinnerton-Dyer Conjecture, I encourage you to watch the video below.

8- The Kissing Number Problem

The Kissing Number Problem has stumped mathematicians for centuries. The problem involves finding the maximum number of equal-sized spheres that can touch one central sphere without overlapping or leaving any spaces between them (Source: Princeton University )

Initially thought to be a simple problem to solve, it is quite challenging to determine this ‘kissing number accurately.’ The answer varies depending on the dimension of the space – in two dimensions, or a flat surface, it’s only six, but in three dimensions, it is much larger and still debated today.

The Kissing Number Problem continues to baffle modern mathematicians, providing an interesting and complex challenge that could lead to countless scientific advances.

Watch the video below to learn more about the Kissing Number Problem.

9- The Unknotting Problem

The Unknotting Problem has fascinated mathematicians since discovering that the unknot is equivalent to a one-dimensional closed loop in three-dimensional space.

The Unknotting Problem simply asks if a particular knot can be undone without changing its form. For example, questions such as “which knots have the fewest crossings?” or “can all knots be unknotted?”. It was first described in 1904 by Greek mathematician Peter Guthrie Tait . It is still an open problem with no known general algorithm for efficiently deciding whether a knot can be untied to become just a circle.

As fascinating as this perplexing problem is to mathematicians, understanding and solving the Unknotting Problem could be essential for researchers hoping to apply mathematics to biology and chemistry, where twists and turns play an important role in the workings of molecules.

Check out A Journey From Elementary to Advanced Mathematics: The Unknotting Problem if you want to learn more.

Here is an interesting PPT presentation about the Unknotting Problem.

What to read next:

• Can You Do A Level Maths In 1 Year? (And How to Ace A Level Math in a Year!)
• What Does a Level Math Course Cover?
• Introduction to Logarithmic Functions .

Wrapping Up

Mathematicians have been tackling difficult mathematical problems since time immemorial with varying degrees of success; some have been solved, while others remain unsolved mysteries today.

From Goldbach’s Conjecture to Fermat’s Last Theorem, challenging mathematical problems continue to captivate mathematicians everywhere and provide them with fascinating puzzles to solve.

Whether you are a high school student studying calculus or a college student taking a higher-level math course, there is always something interesting waiting for you if you look for it.

I encourage you to give some of these hardest math problems a try. Who knows – maybe you will come up with an answer and be rewarded with a million dollars prize.

I am Altiné. I am the guy behind mathodics.com. When I am not teaching math, you can find me reading, running, biking, or doing anything that allows me to enjoy nature's beauty. I hope you find what you are looking for while visiting mathodics.com.

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10 World’s Hardest Math Problems With Solutions and Examples That Will Blow Your Mind

• Updated: August 31, 2023
• Category: Advanced Math , Grade 3 , Grade 4 , Grade 5 , Grade 6

Update : This article was last updated on 12th Oct 2023 to reflect the accuracy and up-to-date information on the page.

The mystical world of mathematics—is home to confounding problems that can make even the most seasoned mathematicians scratch their heads. Yet, it’s also a realm where curiosity and intellect shine the brightest. 

Here are 10 of the world’s hardest math problems , with solutions and examples for those that are solved and a humble “unsolved” tag for the puzzles that continue to confound experts.

1. The Four Color Theorem

Source : Research Outreach

Problem : Can every map be colored with just four colors so that no two adjacent regions have the same color?

Status : Solved

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.

2. Fermat’s Last Theorem

Problem : There are no three positive integers a,b,c that satisfies

a n +b n =c n for n>2.

Solution Example : Andrew Wiles provided a proof in 1994. To understand it, one would need a deep understanding of elliptic curves and modular forms. The proof shows that no such integers a,b,c can exist for n>2.

3. The Monty Hall Problem

Source:  Towards Data Science

Problem : You’re on a game show with three doors. One hides a car, the others goats. After choosing a door, the host reveals a goat behind another door. Do you switch?

Solution Example: Always switch. When you first choose, there’s a 1/3 chance of picking the car. After a goat is revealed, switching gives you a 2/3 chance of winning. If you don’t believe it, try simulating the game multiple times.

4. The Travelling Salesman Problem

Source : Brilliant

Problem: What’s the shortest possible route that visits each city exactly once and returns to the origin?

Status: Unsolved for a general algorithm

Solution Example: This is known as computer science’s most well-known optimization problems. Although there is no solution for all cases, algorithms like the Nearest Neighbor and Dynamic Programming can provide good approximations for specific instances.

5. The Twin Prime Conjecture

Source: Hugin

Problem : Are there infinitely many prime numbers that differ by 2?

Status : Unsolved

Solution Example : N/A

Recommended Reading: Pros and Cons of Math Competition

6. The Poincaré Conjecture

Problem : Can every simply connected, closed 3-manifold be homomorphic to the 3-sphere?

Solution Example : Grigori Perelman proved this using Richard Hamilton’s Ricci flow program. In simple terms, he showed that every shape meeting the problem’s criteria can be stretched and shaped into a 3-sphere.

7. The Goldbach Conjecture

Source: Medium

Problem : Can every even integer greater than 2 be expressed as the sum of two prime numbers ?

8. The Riemann Hypothesis

Source:  The Aperiodical

Problem : Do all non-trivial zeros of the Riemann zeta function have their real parts equal to 1/2?

9. The Collatz Conjecture

Source: Python in Plain English

Problem : Starting with any positive integer n, the sequence n,n/2,3n+1,… eventually reaches 1.

Status: Unsolved

Solution Example: N/A

10. Navier–Stokes Existence and Smoothness

Problem: Do solutions to the Navier–Stokes equations exist, and are they smooth?

There you have it—10 of the world’s hardest math problems. Some have been gloriously solved, giving us a brilliant glimpse into the capabilities of human intellect. Others still taunt the academic world with their complexity. For math lovers, this is the playground that never gets old, the arena where they can continually hone their problem-solving skills. So, do you feel up to the challenge?

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does solving hard math questions even sharpen your brain? I mean I get exhausted and start panicking

Yes, solving tough math problems sharpens your brain by boosting critical thinking and problem-solving skills. While it may feel challenging, the process enhances cognitive abilities over time.

Don’t panic, take breaks, have some time, you’ll get better eventually, not immediately.

Do you know there is an award price of $1 Million for solving the Riemann Hypothesis….. THIS IS SOMETHING SOO SERIOUS!!!!

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Problems & Exercises

1.2 physical quantities and units.

The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?

A car is traveling at a speed of 33 m/s 33 m/s size 12{"33"" m/s"} {} . (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h 90 km/h size 12{"90"" km/h"} {} speed limit?

Show that 1 . 0 m/s = 3 . 6 km/h 1 . 0 m/s = 3 . 6 km/h size 12{1 "." 0`"m/s"=3 "." "6 km/h"} {} . Hint: Show the explicit steps involved in converting 1 . 0 m/s = 3 . 6 km/h. 1 . 0 m/s = 3 . 6 km/h. size 12{1 "." 0`"m/s"=3 "." "6 km/h"} {}

American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)

What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

The speed of sound is measured to be 342 m/s 342 m/s size 12{"342"" m/s"} {} on a certain day. What is this in km/h?

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years?

(a) Refer to Table 1.3 to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second. (b) What is this in meters per second?

1.3 Accuracy, Precision, and Significant Figures

Express your answers to problems in this section to the correct number of significant figures and proper units.

Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

(a) A car speedometer has a 5.0 % 5.0 % size 12{5.0%} {} uncertainty. What is the range of possible speeds when it reads 90 km/h 90 km/h size 12{"90"" km/h"} {} ? (b) Convert this range to miles per hour. 1 km = 0.6214 mi 1 km = 0.6214 mi size 12{"1 km" "=" "0.6214 mi"} {}

An infant’s pulse rate is measured to be 130 ± 5 130 ± 5 size 12{"130" +- 5} {} beats/min. What is the percent uncertainty in this measurement?

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?

A can contains 375 mL of soda. How much is left after 308 mL is removed?

State how many significant figures are proper in the results of the following calculations: (a) 106 . 7 98 . 2 / 46 . 210 1 . 01 106 . 7 98 . 2 / 46 . 210 1 . 01 size 12{ left ("106" "." 7 right ) left ("98" "." 2 right )/ left ("46" "." "210" right ) left (1 "." "01" right )} {} (b) 18 . 7 2 18 . 7 2 size 12{ left ("18" "." 7 right ) rSup { size 8{2} } } {} (c) 1 . 60 × 10 − 19 3712 1 . 60 × 10 − 19 3712 size 12{ left (1 "." "60" times "10" rSup { size 8{ - "19"} } right ) left ("3712" right )} {} .

(a) How many significant figures are in the numbers 99 and 100? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

(a) If your speedometer has an uncertainty of 2 . 0 km/h 2 . 0 km/h size 12{2 "." 0" km/h"} {} at a speed of 90 km/h 90 km/h size 12{"90"" km/h"} {} , what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h 60 km/h size 12{"60"" km/h"} {} , what is the range of speeds you could be going?

(a) A person’s blood pressure is measured to be 120 ± 2 mm Hg 120 ± 2 mm Hg size 12{"120" +- 2" mm Hg"} {} . What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg 80 mm Hg size 12{"80"" mm Hg"} {} ?

A person measures his or her heart rate by counting the number of beats in 30 s 30 s size 12{"30"" s"} {} . If 40 ± 1 40 ± 1 size 12{"40" +- 1} {} beats are counted in 30 . 0 ± 0 . 5 s 30 . 0 ± 0 . 5 s size 12{"30" "." 0 +- 0 "." 5" s"} {} , what is the heart rate and its uncertainty in beats per minute?

What is the area of a circle 3 . 102 cm 3 . 102 cm size 12{3 "." "102"" cm"} {} in diameter?

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

A marathon runner completes a 42 . 188 -km 42 . 188 -km size 12{"42" "." "188""-km"} {} course in 2 h 2 h size 12{2" h"} {} , 30 min, and 12 s 12 s size 12{"12"" s"} {} . There is an uncertainty of 25 m 25 m size 12{"25"" m"} {} in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?

The sides of a small rectangular box are measured to be 1 . 80 ± 0 . 01 cm 1 . 80 ± 0 . 01 cm size 12{1 "." "80" +- 0 "." "01"" cm"} {} , {} 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm size 12{2 "." "05" +- 0 "." "02"" cm, and 3" "." 1 +- 0 "." "1 cm"} {} long. Calculate its volume and uncertainty in cubic centimeters.

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm = 0 . 4539 kg 1 lbm = 0 . 4539 kg size 12{1" lbm"=0 "." "4539"`"kg"} {} . (a) If there is an uncertainty of 0 . 0001 kg 0 . 0001 kg size 12{0 "." "0001"`"kg"} {} in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

The length and width of a rectangular room are measured to be 3 . 955 ± 0 . 005 m 3 . 955 ± 0 . 005 m size 12{3 "." "955" +- 0 "." "005"" m"} {} and 3 . 050 ± 0 . 005 m 3 . 050 ± 0 . 005 m size 12{3 "." "050" +- 0 "." "005"" m"} {} . Calculate the area of the room and its uncertainty in square meters.

A car engine moves a piston with a circular cross section of 7 . 500 ± 0 . 002 cm 7 . 500 ± 0 . 002 cm size 12{7 "." "500" +- 0 "." "002"`"cm"} {} diameter a distance of 3 . 250 ± 0 . 001 cm 3 . 250 ± 0 . 001 cm size 12{3 "." "250" +- 0 "." "001"`"cm"} {} to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.

1.4 Approximation

How many heartbeats are there in a lifetime?

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10 − 22  s 10 − 22  s size 12{"10" rSup { size 8{ - "22"} } " s"} {} .)

Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10 − 27  kg 10 − 27  kg size 12{"10" rSup { size 8{ - "27"} } " kg"} {} and the mass of a bacterium is on the order of 10 − 15  kg. 10 − 15  kg. size 12{"10" rSup { size 8{ - "15"} } "kg"} {} )

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

(a) What fraction of Earth’s diameter is the greatest ocean depth? (b) The greatest mountain height?

(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?

Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-physics/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

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• Publisher/website: OpenStax
• Book title: College Physics
• Publication date: Jun 21, 2012
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10 Hardest AP Chemistry Exam Questions

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How will ap scores impact my college chances, overview of the ap chemistry exam, 10 hardest ap chemistry questions.

In 2020, 56.1% of test takers passed the AP Chemistry exam with a 3 or higher. As you can see, the AP Chemistry exam is one of the more difficult exams administered by the College Board. But, with the right study tools, such as CollegeVine’s Ultimate Guide to the AP Chemistry Exam , and through lots of practice, you will be more than prepared to pass the exam! 

In this article, we will review some of the more difficult questions you may encounter on the exam followed by detailed explanations describing how to solve them.

The score you get on an AP exam, whether it be a 2 or a 5, won’t really impact your chances of admission into a particular college. Most colleges will not even review your scores , especially since you don’t have to send them in if you don’t want to!

College admissions will instead take a more holistic approach when reviewing your college application. Course rigor, GPA,  essays, and extracurriculars are more likely to be valued when reviewing your application. When it comes to course rigor, colleges want to see how many APs you take and what grades you get in the classes rather than the exam scores, which are used to determine course credit and placement after you get into college. 

If you are interested in understanding how your course load impacts your chances at college, use our free chancing engine . This tool will help you determine your chances of acceptance at over 500 colleges based on your grades, test scores, extracurriculars and demographics. Keep in mind that your essays and letters of recommendation are also important in the college admissions process!

The AP Chemistry exam assesses your ability to understand science practices in relation to four big ideas: scale, proportion and quantity; structure and properties; transformations; and energy. The exam is 3 hours and 15 minutes long and is comprised of two separate sections: 

Section I: Multiple Choice Questions

• 60 questions
• A calculator is permitted
• Worth 50% of your exam score

Section II: Free Response Questions(FRQs)

• 3 Long-Answer Questions
• 4 Short-Answer Questions
• 105 minutes
• A calculator is NOT permitted

The AP Chemistry exam is broken up into nine units with the following percentage breakdowns:

• Atomic Structure and Properties (7-9%)
• Molecular and Ionic Compound Structure and Properties (7-9%)
• Intermolecular Forces and Properties (18-22%)
• Chemical Reactions (7-9%)
• Kinetics (7-9%)
• Thermodynamics (7-9%)
• Equilibrium (7-9%)
• Acids and Bases (11-15%)
• Applications of Thermodynamics (7-9%)

In this article, we will be focusing on official CollegeBoard AP Chemistry Exam questions. This means that you can expect to see the same level of difficulty on the exam you take. 

CollegeBoard is also known to provide answer options that are meant to confuse you into choosing the wrong answer instead of the correct one. But fear not! We will show you how to carefully read the question and make sure that you are confident in the answer you choose!

Answer: (A)

In this first question, we see that a base Mg(OH) 2 is being titrated by an acid HNO 3 . In order to determine the mass of the initial base, we need to perform a series of titration calculations. This is a great example of when your labs come in handy, since the lab can help you visualize what is going on. 

First, we need to determine the reaction that is occurring when Mg(OH) 2   is reacted with HNO 3 . We first write out the equation we know, which is that the base and acid react to create water and an aqueous solution of the remaining ions:

Mg(OH) 2   + HNO 3 → H 2 O + Mg 2 + + NO 3 –

We now need to balance the above reaction. Hydrogen is already balanced, so the next element we move to is Oxygen. We see that Oxygen is unbalanced with five oxygen on the left and four oxygen on the right. We can balance Oxygen by:

Mg(OH) 2   + 2HNO 3 → 2H 2 O + Mg 2 + + 2NO 3 –

This gives us 8 oxygen molecules on either side. This simple balancing act has now balanced the entire equation with one Mg molecule, 8 O molecules, 4 H molecules and 2 N molecules on either side. 

Now we return to the titration problem. Since we know that the acid is being used as the titration solution, we know that HNO 3 is in the buret. We also know that 20 mL is needed to get to the end point since:

22.17 mL – 2.17 mL = 20 mL

Now that we know the molar mass of the base (58.3 grams), the molarity of the acid solution, and the amount of acid needed to titrate the bass and the balanced equation, we can perform a simple stoichiometric equation. Make sure to balance all your units and note that it takes 2 moles of HNO 3 for every one mole of Mg(OH) 2 !

Answer: (E)

Geometry of a molecule or compound is an important topic covered on the AP Chemistry exam so it is important that you know your tetrahedrals from your trigonal bipyramidials! An easy hint is that you can almost always determine what type of shape a compound is based on the number of lone electron pairs and molecules that are attached to the central atom.

That being said, a planar compound is one where the atoms surrounding the central molecule arrange themselves so that they exist on a singular, two-dimensional plane. In order to determine which of these is NOT planar (remember to read extra carefully!!!), we’ll draw out each of the molecules. Make sure to follow the quartet rule when drawing out each of the compounds!

This carbonate ion, while experiencing negative charges on the oxygen, is planar since all oxygens are in the same plane as one another.

The nitrate ion is also planar since the oxygen atoms are in the same plane as one another and are not impacted by lone pairs (similar to the Carbonate ion).

Now this is a tricky compound. At first glance, you would assume that ClF 3 is not planar due to the impact of the two electron pairs present. However, the chlorine trifluoride compound is actually a T-shaped geometrical structure. In this shape, the 3 Fluorine atoms are in the same plane and form a T shape with the Chlorine atom. Surprisingly, this is not the correct answer.

This compound is another simple answer we can cross off since the three Fluorine atoms are in the same plane as one another.

This leaves us with phosphorus trichloride. The electron pair causes a disruption in what would be expected of a planar compound. Each individual chlorine atom experiences great repulsion from the lone electron pair, thereby causing the Chlorine atoms to bend closer together. This gives this compound a trigonal pyramidal shape instead of a planar geometry.

Therefore, the compound that is not planar is (E) PCl 3 .

This is a tricky question since we are given the initial mass of both reactants and are asked to find the maximum mass of a product, in this case water. This means this is a limiting reactant problem in which a limiting reactant limits how much a product can be produced. 

Let’s perform stoichiometric equations on 8 grams of N 2 H 4 as well as 92 grams of N 2 O 4 to determine which of these reactants is the limiting reactant. But first, always, always, always double check that the provided equation is balanced! (Thankfully in our case, it is).

Here we see that while N 2 O 4 may produce the most amount of water, this is an inaccurate representation of what happens. N 2 H 4 acts as a limiting reactant and limits how much water is produced. N 2 O 4 is in excess. Therefore, the maximum amount of water produced is 9 grams.

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Answer: (D)

This question is difficult since it’s not a problem that is often seen in class. The main thing we need to understand is that regardless of how difficult this question looks, ultimately, it’s another balancing problem!

In this reaction, neutrons are being used to induce a fission reaction. A neutron is the same mass as a proton, however, has no charge (no electrons or protons).

Remember that the upper number next to the element represents the atomic mass while the bottom number represents the atomic number. 

To simplify the reaction we can expand the equation to account for the three neutrons that are being displaced:

92 235 U + 1 0 n → 55 141 Cs +  0 1 n + 0 1 n  + 0 1 n  + ? ? X

This means that we can perform simple subtraction to determine what ? ? X  is since the law of conservation of mass must be obeyed:

(92)-(55) (235+1) – (141+1+1+1) X = 37 92 X

By checking a periodic table of elements, we see that Rb has an atomic mass of 92 and an atomic number of 37.

Answer: (B)

This is another EXCEPT question so we have to be extra careful as to what is being asked of us! Remember that a Lewis acid-base reaction is one in which a Lewis-base donates it’s extra pair of electrons to a Lewis acid.

In the five options above, OH – , CL – , NH 2 – and NH 3 have electron pairs that are capable of donating their additional electrons to the Lewis acid AL(OH) 3 , SnCl 4 , NH 4 + and H + respectively. In option (b), H 2 O acts as a nucleophile and binds with Cl 2 and breaks the intramolecular Cl 2 bond. While doing so, the additional hydrogen atom breaks off in order to stay true to the octet rule. In doing so, the reaction fails to create a complete covalent bond between the two molecules.

This question is difficult because it requires you to understand how a buffer and titration works and know how to approximate data based on a graph.

The optimum buffer action occurs when the amounts of acid and its conjugate base are approximately equal. In the case of our situation, this happens when half of the weak acid has been neutralized to form its conjugate base. 

It may seem like the best answer for this question is point W, the point when there is enough base to neutralize the acid (also known as the equivalence point). But this is incorrect! This point basically neutralizes the entire solution and that’s not what we’re looking for! At this point, the strong base is in excess in relation to the weak acid to guarantee that all of the weak acid is neutralized (as seen by the pH value of 8.0 at point W).

We are looking for the point when the buffer action is optimal, which as we defined before, is when there are equal amounts of acid and its conjugate base. This point occurs approximately midway between the equivalence point (at which the slope of the graph is nearly infinite) and the starting point. In relation to the graph this occurs at point V.

Balancing equations always proves to be time consuming and difficult, especially when performed under pressure. However, it is almost guaranteed that these types of problems will appear on the exam. 

This type of question is best solved as reduction and oxidation half reactions To do this, we need to determine which elements are being reduced and which elements are being oxidized. We see that the chromium ion gains electrons, therefore becoming reduced, while sulfur loses electrons when it becomes a solid, therefore becoming oxidized. 

Reduction: Cr 2 O 7 – (aq) + H + (aq) → Cr 3+ (aq) + H 2 O(l)

Oxidation: H 2 S(g) → S(s)

Next, we must balance all the atoms in the above equation. We can account for missing Hydrogen ions by adding in protons:

Reduction: Cr 2 O 7 – (aq) + 14H + (aq) → 2Cr 3+ (aq) + 7H 2 O(l)

Oxidation: H 2 S(g) → S(s) + 2H +  

From here, we must determine how many electrons are needed to balance out the charges of either side, which we can do by adding 6 electrons to the left side of the reduction equation and 2 electrons the the right side of the oxidation equation:

Reduction: Cr 2 O 7 – (aq) + 14H + (aq) + 6e – → Cr 3+ (aq) + 7H 2 O (l)

Oxidation: H 2 S(g) → S(s) + 2H +  + 2e –  

The next goal is to cancel out the electrons from either side. We can do this by multiplying the oxidation equation by 3. This then becomes:

Oxidation: 3H 2 S (g) → 3S(s) + 6H + + 6e –

Now we can add the respective equations to get:

Cr 2 O 7 – (aq) + 14H + (aq) + 3H 2 S (g) +  6e –    → Cr 3+ (aq) + 7H 2 O (l) + 3S(s) + 6H +   + 6e –

The last step is to cancel out any like terms on opposite sides:

Cr 2 O 7 – (aq) + 8H + (aq) + 3H 2 S (g)   → Cr 3+ (aq) + 7H 2 O (l) + 3S(s)

Therefore, the coefficient of H +  is 8!

This question requires us to remember concepts of solubility constant as well as how reactions work in order to conserve mass. The first thing we must do is to write out the reaction that is described in the problem statement. The thing that makes this question difficult is writing out the correct reaction. Why you may ask? It’s because the correct equation involves balancing out the charges of the atoms!

Zn(OH) 2 → 2OH – + Zn 2+

We know that K sp is written as the product of the concentrations of the aqueous ions in a solution. Since OH – has a coefficient of 2, we must square the concentration of the base. Therefore:

K sp = [OH – ] 2 [Zn 2+ ]

From here, we are dealing with easy math! We know that the hydroxide ion and Zinc need to dissolve in proportional concentrations. Therefore, we can reduce the equation in terms of x . But, don’t forget to include the coefficient of 2 in front of the hydroxide ion as well!

K sp = [2x] 2 [x]

We are given the concentration of the hydroxide ion as 2.0 x 10 -6 M. This means that:

2x = 2.0 x 10 -6 M

x = 1.0 x 10 -6 M

Plugging these values into the above equation gives us:

K sp = [2.0  x 10 -6 ] 2 [1.0 x 10 -6 ] = 4.0 x 10 -18 M

Answer: (C)

This type of question seems difficult to look at since you are immediately presented with numerous compounds you may have not seen before. With this type of question, it is important to remember characteristics about specific compounds such as halides and sulfates.

One fact to remember about carbonates (CO 3 2- ) is that they are most often insoluble. They are especially insoluble when paired with Group II atoms such as Barium, Calcium and Strontium. Therefore, BaCO 3 is the least soluble in water of the groupings.

Question 10

This is another one of those questions where it is extremely helpful if you are able to remember labs shown or conducted in your class. This question also requires you to remember qualitative characteristics of compounds to deduce what ions could be present in an unknown solution. 

The initial white precipitate formed corresponds to a chloride precipitate since hydrochloric acid was used to treat the unknown solution. However, we know that BaCl 2 is soluble, so we can deduce that Barium is not in the initial solution since a solid forms after the treatment. 

This leaves us with AgCl, HgCl 2 and PbCl 2 as potential precipaties. Since a bright yellow precipitate formed after the introduction of the potassium iodide solution, we know that this is a characteristic of lead iodide. Therefore, lead is present in the solution. This leaves us with answer choice (B) as the right answer.

To verify that B is correct, we must determine that silver is also present in the unknown solution. We know this to be true because silver easily dissolves in ammonia, as described in the problem statement.

As you prepare for this exam, remember to:

Know Your Labs!

Many of the multiple choice questions and free response questions you’ll come across will describe lab scenarios. These scenarios may have been ones you have previously seen in class or in a video. It’ll be easier for you to understand and correctly answer the question at hand if you are able to visualize what is being described based on something you have seen previously in class!

Practice, Practice, Practice!

You may know the concepts very well, but you also need to be able to apply the concepts in a timed setting. Make sure you take multiple practice tests administered through CollegeBoard or through your school. You might be able to answer every question on the test perfectly, but you need to be sure that you can finish the multiple choice section in 90 minutes! Remember, that leaves just about one and a half minutes to read the question and determine the answer. Over time, you will get better at quickly answering questions and determining what the question is asking for on your first try! 

The Night Before…

Review your important equations! Know what each equation on the formula sheet means. It’s one thing to know what the variables on the equation sheet mean, but it’s another to know why you would use an equation and how it can help you solve any question at hand. 

People will always tell you to get a good eight hours of sleep and eat a healthy breakfast. While this may seem like basic info that is not relevant to the test material you are memorizing, it is important that you are awake and ready to sit through a three hour exam!

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Demystifying Math 55

By anastasia yefremova.

Few undergraduate level classes have the distinction of nation-wide recognition that Harvard University’s Math 55 has. Officially comprised of Mathematics 55A “Studies in Algebra and Group Theory” and Mathematics 55B “Studies in Real and Complex Analysis,” it is technically an introductory level course. It is also a veritable legend among high schoolers and college students alike, renowned as — allegedly — the hardest undergraduate math class in the country. It has been mentioned in books and articles, has its own Wikipedia page, and has been the subject of countless social media posts and videos.

Most recently, Harvard junior Mahad Khan created a TikTok video dedicated to Math 55 that has received over 360,000 views to date. His is only one of many — his older brother created one, too — but it has the distinction of an insider’s perspective. “I thought it would be interesting if I cleared up the misconceptions about Math 55,” Khan said. While he hadn’t taken the course himself, he wanted to go beyond its reputation. “I wanted to get a real perspective by interviewing a former student and current course assistant.”

Over the years, perception of Math 55 has become based less on the reality of the course itself and more on a cumulative collection of lore and somewhat sensationalist rumors. It’s tempting to get swept up in the thrill of hearsay but while there might be kernels of truth to some of the stories, many of them are outdated and taken out of context. At the end of the day, however, Math 55 is a class like any other. Below, we take a stab at busting some of the more well known and persistent myths about the class. Or, at the very least, offering an extra layer of clarity. 

Myth #1: Math 55 is only for high school math geniuses

Most articles or mentions of Math 55 refer to it as filled with math competition champions and genius-level wunderkinds. The class is supposedly legendary among high school math prodigies, who hear terrifying stories about it in their computer camps and at the International Math Olympiad. There are even rumors of a special test students have to take before they are even allowed into Math 55. But while familiarity with proof-based mathematics is considered a plus for those interested in the course, there is no prerequisite for competition or research experience. 

In fact students whose only exposure to advanced math has been through olympiads and summer research programs can have a harder time adjusting. Their approach to the material tends to be understandably more solitary and that can be a disadvantage for the level of collaboration higher level mathematics require. “It has become a lot more open to people with different backgrounds,” said Professor Denis Auroux , who teaches Math 55,. “Our slogan is, if you’re reasonably good at math, you love it, and you have lots of time to devote to it, then Math 55 is completely fine for you.” 

Also, there is no extra test to get into the class.

Myth #2: Just take a graduate class, instead

Math 55 is hard. Whether you’re just 55-curious, or a past or present student in the class, this is something everyone agrees on. The course condenses four years of math into two semesters, after all. “For the first semester, you work on linear and abstract algebra with a bit of representation theory,” said sophomore math concentrator Dora Woodruff. “The second semester is real and complex analysis, and a little bit of algebraic topology. That’s almost the whole undergraduate curriculum.” Woodruff — incidentally, the student Khan interviewed — took Math 55 as a freshman and returned her second year as a course assistant. She is intimately familiar with the course’s difficulty level.

So why not just take an upper level undergraduate course to begin with or even one at a graduate level, if you’re really looking for a challenge? What justifies the existence of a class with the difficulty level of Math 55? One argument is that the course helps structure and systemize the knowledge with which many students come to Harvard. It gives them a firm background in preparation for the rest of their math education. Math 55 is difficult and it is purposefully structured that way as it’s meant to help students mature as mathematicians rather than as simple course takers.

But more importantly, “it’s just not true that Math 55 is at the level of a graduate class,” Auroux said. “It goes through several upper division undergraduate math classes with maybe a bit more advanced digressions into material here and there, but it sticks very close to what is taught in 100-level classes. The difference is we go through it at a faster pace, maybe with more challenging homework, and ideally as a community of people bringing our heads together.” 

A core goal of Math 55, according to Auroux, is to build a sense of community. Other schools might encourage advanced first-year students to take upper level undergraduate or even graduate classes, but Math 55 helps build a cohort of like-minded people who really like math, are good at it, and want to do a lot of it during their time at Harvard. That’s the experience Woodruff had, as well. “The community can be very strong,” she said. “You meet a lot of other people very interested in math and stay friends with them for the rest of college.”

Myth #3: Homework takes between 24 and 60 hours

Horror stories of endless homework are synonymous with the class. You’ll read or hear about “24 to 60 hours per week on homework” in almost every reference to Math 55. But one, there is a world of difference between 24 and 60 hours that is never explained, and two, this timeframe is quite misaligned with reality.

Auroux frequently sends out surveys to his students asking how long homework takes them and the average for most is closer to 15 hours a week. Those with more extensive prior math backgrounds can take as little as five to ten hours. The key factor is collaboration. “This class doesn’t lend itself to self-study,” Auroux stressed. Once they have thought about each problem set on their own, students are welcome and encouraged to talk to their friends and collaborate. “As soon as I see that something took over 30 hours I ask the student, do you know you’re supposed to be working with people and come ask me questions when you’re stuck?”

It is true that between reviewing lectures, digesting the material, and solving the problem sets, students usually end up devoting between 20 and 30 hours a week to the class. However, that includes the time dedicated to homework. So while students are discouraged from taking too many difficult classes and extracurriculars in the same semester as Math 55, they are also not expected to spend the time equivalent to a full-time job on their problem sets every week.

Myth #4: less than half of the class makes it to the second semester

Math 55 is just as infamous for its attrition rate as it is for its difficulty. Most sources like to cite the 1970 class, which began with 75 students and — between the advanced nature of the material and the time-constraints under which students had to work — ended with barely 20. Since then, the rumor has been that the Math 55 class shrinks by half its original size or more before the first semester is over. The reality is much less shocking and a bit more complicated.

Enrollment in this past fall semester’s Math 55A peaked at (ironically) 55 students. Well into the spring semester’s Math 55B, 47 students were still enrolled in the course. “On average, a drop of about 10-15 percent is much closer to what I would expect,” Auroux said. And those numbers become even more flexible if one takes into consideration the weeks math students have at the beginning of each semester to try out different classes and “shop” around before they have to commit to anything. This means students find their way in and out of Math 55 in a variety of ways over the course of the academic year.

According to Auroux, some students shop Math 55 in the fall and switch to the less intense Math 25 for the remainder of the semester. Others start out in Math 25 and, if not sufficiently challenged, switch to Math 55. Even people who end up in academia are not exempt from this. During his time as a student, our own Department of Mathematics’ Professor Emeritus Benedict Gross switched to the lower level Math 21 after two weeks in Math 55. In fact, those two weeks almost made him reconsider his desire to pursue mathematics. “By the beginning of sophomore year, I had decided to major in physics,” he recalled. “But during shopping period that fall, I walked past a math class taught by Andrew Gleason and stopped in to listen. It turned out to be Math 55.” He enrolled and by the end of the semester had found his vocation in mathematics.

All this means that Auroux sees student numbers vacillate up and down throughout the academic year. “There are about four or five students in this spring semester’s Math 55 that took Math 25 or even Math 22 in the fall, and they’re doing mostly fine,” he said. “It’s a lot of work, but I think they’re having a great time.”

Myth #5: 55-er culture is cult-y and exclusionary

Even though her experience with Math 55 was a positive one, Woodruff is very aware of the unhealthy culture the class has been rumored to cultivate. It’s easy for students to form exclusionary cliques that consist only of other Math 55 students, and some look down on anyone taking lower level math classes. But Woodruff also stressed that the instructors are very aware of this and actively take steps to curb that kind of toxic behavior. She said Auroux frequently brings up the importance of keeping the Math 55 community inclusive through Slack messages and lecture references.

Some students come to Harvard just for the opportunity to take Math 55. Some view enrolling in the class as proof of their mathematical gumption and competence. A Harvard Independent article called Math 55 the “premiere mathematical challenge for overachieving and…ridiculously mathy freshmen” and a piece in The Harvard Crimson referred to it as “a bit of a status thing as far as math majors here are concerned.” Over the years, the Harvard Department of Mathematics has taken steps to correct these assumptions. 

For one thing, neither the Math 55A nor the Math 55B official course descriptions boast the dubious honor of referring to it as “probably the most difficult undergraduate math class in the country” (don’t trust everything you read on Wikipedia). For another, “we’re trying to emphasize that there’s no magic to Math 55,” Auroux said. “It contains the same material as some of the other classes we have. People who take it are not intrinsically better or smarter than the ones who don’t.” 

Myth #6: You have to take Math 55 if you’re serious about going into academia

One reason math concentrators could feel pressured to enroll in Math 55 is because they view it as a prerequisite for a career in academia. It’s a sort of badge of honor and proof of their commitment to the field of mathematics. It is true that quite a few graduates of the course have gone on to pursue a career in mathematics. Woodruff herself believes that will be the most likely path for her, and several faculty members in our own Department of Mathematics took Math 55 during their days as Harvard freshmen.

“Several times in my research career when I understood something fundamental, I would realize that this was what Math 55 was trying to teach us,” Gross said. “It was an amazing introduction to the whole of mathematics and it was transformative for me.” In fact, Gross met Higgins Professor of Mathematics Joe Harris when they took the class together, forging a lifelong friendship. When they returned to Harvard as faculty, they took turns teaching Math 25 and Math 55. 

However, Auroux is quick to point out that while many graduates of the course do end up in academia, most professional mathematicians have likely never even heard of Math 55. “I would like to think that it’s a success story if people end up doing math, because the goal of Math 55 is to show students how beautiful math can be,” he said. “If they love it enough to go to grad school and become mathematicians, that’s wonderful. And if they want to take that math knowledge and do something else with their life, that’s just as wonderful.”

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Test Yourself: The Difficult Homework Questions That Parents Can’t Even Answer

Reporter at HuffPost UK

Last time we checked, homework is meant to consolidate what children are learning at school , not test the brains of parents who are several decades older (and hopefully a little wiser).

But time and time again parents are taking to the internet to share the utterly baffling work designed for kids, which is leaving adults feeling outwitted.

Can you answer these homework questions that other families are struggling to get to the bottom of?

1. This numbers question.

The Holderness family uploaded this photo of their son’s complex homework on Facebook with the caption: “Internet friends: solve this 1st grade math homework # showyourwork # mybrainhurts.”

Try it for yourself here.

2. This shapes question.

This dad was helping his son , who is in year five, at a primary school in Glossop, Derbyshire, when they reached this question in his math’s homework.

The 43-year-old said he “spent an hour” trying to work it out and sending it to friends with economics degrees, who also couldn’t answer it.

3. This fruit-based question.

This is driving Facebook crazy today ... pic.twitter.com/xGd9a7cRLZ — ♛ Melanism •••€ (@_aquemini) January 31, 2016

This coconut conundrum sent Facebook into overdrive last year, after mum Thighler Perry tweeted a picture of the brainteaser.

The task is to work out the missing value on the fourth sum, by finding the values of an apple, banana and coconut. But if you break it down, it’s really not that hard. Or is it?

4. This vocabulary question.

2nd Grade Exam

This teacher marked two answers wrong on the sheet , for a second grade pupil, who got the definitions for ‘twinkle’ and ‘sparkle’ mixed up. Apparently there is difference between what these words mean. Who knew?

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