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Math is often intimidating to the average man due to its complex appearance. To show that math requires creative thinking, not just memorization, I was wondering if anyone had any math problems that required some out-of-the-box thinking.

I am looking for something that can be solved with just basic math skills. Please limit problems that can be solved with high school math. I do not want a problem with a complex solution, just one that requires a lot of thinking to solve.

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Look up Raymond Smullyan's books with a lot of logic puzzles that require very little of what most people consider "math", but which are entirely mathematical. Also, the "census-taker problem" is a nice problem. –  Arturo Magidin May 24 '12 at 22:13
@ArturoMagidin I'll look into that. Thanks! –  user26649 May 25 '12 at 2:33
Math can be intimidating to the average woman, too. –  Potato May 27 '12 at 3:02
I believe old IMO questions are available online. Many are solvable with high school maths. All require a lot of thinking. –  user1729 Jun 10 '12 at 20:43
12 barrels of diesel, each containing 50 gallons, and an empty truck with an empty fuel tank of 10 gallons capacity are available at a base station. the truck can carry only one barrel at a time, and travels exactly 10 mi/gallon diesel can be siphoned at will from barrel-barrel, barrel to truck & vice versa, and barrels (full/partly full/empty) can be left anywhere on a long, lonely road through the desert. all the barrels are at base camp and only this truck is available What is the maximum distance from base camp the truck can travel ? –  true blue anil Jun 13 '12 at 6:59

13 Answers 13

Do there exist infinitely many primes?

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I think the fly and two trains problem is a good example of a creative solution accessible to those with basic math skills.

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There's a creative solution? I just summed the series. –  MJD May 27 '12 at 1:01
The creative solution is in the article. –  Zev Chonoles May 27 '12 at 1:31
That was a joke. That remark is commonly attributed to John Von Neumann. –  MJD May 27 '12 at 1:35

What I think of as the "standard examples" in this area:

Since I don't like the Wikipedia entry for the second thing as much: the basic problem is: given a (connected) graph, to determine whether or not there is a path in the graph that goes over each edge exactly once (what is now called an Eulerian path). In a small graph, if such a path exists, one can easily be found by trial and error (even very young children can do it). But when one cannot find such a path by trial and error, at first glance, it is very difficult to understand (let alone prove) that there aren't any.

Euler's insight was that the answer depends only on the degrees of the vertices in the graph: if a graph has more than two vertices of odd degree, it cannot have such a path. And if you think about it a while, this becomes "obvious", although it is far from obvious if your only experience with the problem is using pen and paper to look for Eulerian paths.

(I should say: it's significantly harder to show that that if a connected graph has two or fewer vertices of odd degree, then it must have an Eulerian path. But with any graph that is small enough to draw in a short amount of time, the truth of this statement can at least be checked by hand.)

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This question is much too vague and too vast to admit a meaningful list of answers. (What are basic math skills? What is a complex solution?)

I suggest that you look at the site


where you can find nice problems of different difficulties that only use high school mathematics.

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How about calculating the sum of the first n integers, and (the legend of) Gauss' solution?

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This was also my first thought. This is also a good example of why a graphical/geometrical/visual intuition is a good tool! –  utdiscant Jun 12 '12 at 21:09

I think the proof of Cantor's Theorem is simple enough to show the creativity found in mathematical arguments without being too deep.

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Say I have a square. I pick 3 random points on the perimeter of the square, each on a different side. What are the odds that the resulting triangle contains the centre of the square?

Can be solved (possibly sacrificing a little rigor) without a single line of calculation.

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Would it be $\frac{1}{4}$? The first point's position is irrelevant. For the triangle to contain the center point the two other points must be below the squares center. Since there are two possibilities for a point's position - above the center and below it, the probability of a point being below the center is $\frac{1}{2}$. Since both points need to be below the center, the total probability would be $(\frac{1}{2})^2$ or $\frac{1}{4}$. I am disregarding the possibility of the point being on the center because of the probability being so low it's trivial. Is my reasoning correct? –  user26649 Jun 13 '12 at 11:35
not quite, though the reasoning is mostly correct. Both points do not have to be below the centre. If I move G to the bottom corner I can put H anywhere and enclose the centre, so that changes your answer. –  Robert Mastragostino Jun 13 '12 at 15:27

Project Euler has a number of both interesting and challenging problems. The discussion board of solutions that opens up after you solve a problem is also insightful and full of creative solutions: http://projecteuler.net/

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This is more related to computer science and simply programming in many cases. But in many of the problems, a good creative mind is essential. –  utdiscant Jun 12 '12 at 21:12

The social Golfer Problem. Kirkmans Fifteen School Girl problem. The motorcycle Problem. The desert exploration problem. Should Phil chop his big toe off problem ?

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So I found the golfer and Kirkman's, and I've heard of the desert exploration, but googling the other two got no results. Explanation? –  Eric Stucky May 27 '12 at 3:32


The Penguin Book of Curious and Interesting Numbers: Revised Edition (Penguin Press Science) [Paperback] David Wells (Author)

and other books by him.

June 7: Here is a puzzle with a solution by Dirac.

5 sailors are shipwrecked on a South Sea Island, and before they go to sleep they collect a pile of bananas to share out in the morning. One of then wakes up, does not trust the others, so divides the pile into 5 equal parts, with one left over, which he gives to the monkey, and then takes his own part for himself. This process repeats with the other sailors. In the morning, they find the pile divides into 5 equal parts, with one left for the monkey.

Problem: How many bananas were there in the initial pile?

Easy solution by Dirac: Start with $-4$ bananas. With one for the monkey, that makes $-5$ which divides into $5$ equal parts; take away $-1$ for the sailor leaves $-4$, as before. So the process repeats. To get a positive solution, add $5^5$. (No wonder the problem was difficult to solve directly!)

As an exercise for students, you can ask them to generalise the problem, since that is what mathematicians do.

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This might look advanced but trust me, simple reasoning and logic gives you a beautiful answer.

A permutation of first $N$ numbers is said to be good if $x$ and $x+1$ do not occur consecutively. For $N=3$, they are $(1,3,2)$, $(2,1,3)$ and $(3,2,1)$, and the number of such permutations are $3$. Find the number of ways for any $n$.

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There is no greatest natural number (or $\mathbb{N}$ does not have an upper bound in $\mathbb{R}$).

It seems daft but a lot of non-mathematicians have argued with me about this.

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Look up ultrafinitism. I think it is a silly philosophy, but it is perhaps a philosophy that you should think about first, before being flippant about it. (Of course, you should be flippant about it, but you should know why you should be flippant about it. Know thine enemy.) –  user1729 Jun 11 '12 at 13:13
Thank you for the link. –  Paul Slevin Jun 12 '12 at 12:43

$$ \lim \sqrt[n] n = 1$$

Requires fairly creative thinking to apply the Squeeze Theorem. $\sqrt[n] n = 1 + t_n$ where $t_n \ge 0$ for each $n \in \Bbb N$. Then $n = (1 + t_n)^n = 1 + nt_n + \frac 1 2n(n - 1)t_n^2 + ... + t_n^n \implies n \gt \frac 1 2n(n - 1)t_n^2$ $\implies 0 \le t_n \lt \sqrt {\frac 2 {n - 1}} \implies \lim t_n = 0 \implies \lim(1 + t_n) = \lim \sqrt[n] n = 1$

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