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 (firstname.lastname@example.org, CH 614) by November 10th. 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.
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, p1, p2, p3, …, p10, with p10 wrapping around to p1. Then p1+p2+p3, p4+p5+p6, and p7+p8+p9 are each divisible by 3. Therefore p1+p2+…+p9 is divisible by 3, so p10 cannot be divisible by 3. Applying the same argument to p2+p3+p4, p5+p6+p7, and p8+p9+p10 shows p1 is not divisible by 3. Applying the argument repeatedly shows none of the numbers p1, p2, …, p10 is divisible by 3, which is clearly false. Therefore, no such arrangement is possible.