2005 Cyprus Seniors TST/Day 1/Problem 4

Problem

Prove that number 6 divides the product of every three consecutive integer numbers and then show that 240 divides the product $(5\nu+1)(5\nu+3)(5\nu+5)$ , where $\nu$ is any odd number.

Solution

We first prove that there is a multiple of 2 and a multiple of 3 in three consecutive integers:

n, n+1, n+2

If n is odd, n+1 is even, and therefore a multiple of 2. If n is even, n is a multiple of 2.

n can be a multiple of 3. If $n\equiv 1 \pmod 3$ , then n+2 is a multiple of 3. If $n\equiv 2 \pmod 3$ , n+1 is a multiple of 3.

Therefore, there is a multiple of 2 and a multiple of 3 in any three consecutive integers. Therefore, the product of three consecutive integers is a multiple of 6.

Now we prove the second:

$\nu=2x-1$