# 2021 Math Contest: Math Prize Problem #1

Each month the Mathematics Department offers a math prize problem – a mathematical brain teaser – with a first prize of $25, a second prize of $15, and, of course, bragging rights. These problems do not require advanced knowledge of mathematics, just a curious mind and a willingness to slug it out with the problem.

### Arrange the ten integers 1, 2, 3, …, 10 in a circle so that the sum of any three consecutively positioned integers is divisible by 3, or show that it is impossible.

Submit a solution to this problem to Prof. Ken Ching (kching@mmm.edu, CH 614) by **November 10 ^{th}**. The first correct submission will win a

**$25 cash prize**. All other correct submissions will receive recognition and bragging rights, and one of these submissions will be randomly selected for a $15 second prize. All are welcome to participate, but cash prizes are for MMC students only. Solution will be posted online and outside CH 603.

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*We did not have a winner this time!*

*We did not have a winner this time!*

## Solution to 2021 Math Prize Problem #1:

Arrange the integers 1, 2, 3, …, 10 in a circle so that the sum of any three consecutively positioned integers is divisible by 3, or show that it is impossible.

It is impossible. First, note that 1+2+3+…+10 = 55 is not divisible by 3 (it has a remainder of 1 when divided by 3). Suppose such an arrangement is possible, say, p_{1}, p_{2}, p_{3}, …, p_{10}, with p_{10} wrapping around to p_{1}. Then p_{1}+p_{2}+p_{3}, p_{4}+p_{5}+p_{6}, and p_{7}+p_{8}+p_{9} are each divisible by 3. Therefore p_{1}+p_{2}+…+p_{9} is divisible by 3, so p_{10} cannot be divisible by 3. Applying the same argument to p_{2}+p_{3}+p_{4}, p_{5}+p_{6}+p_{7}, and p_{8}+p_{9}+p_{10} shows p_{1} is not divisible by 3. Applying the argument repeatedly shows none of the numbers p_{1}, p_{2}, …, p_{10} is divisible by 3, which is clearly false. Therefore, no such arrangement is possible.

**Published:** October 19, 2021