1. The problem statement, all variables and given/known data show that n(n-1)(2n-1) is always divisible by 6. 3. The attempt at a solution I see how to get the factor of 2 if n is even then n is divisible by 2. is n is odd then n-1 is even. Here is my idea to get the factor of 3. we know that n and n-1 are consecutive integers. and we know that if n is divisible by 3 then so is 2n and the same goes for n-1 but 2n-1 comes right before 2n, so we could some how make an argument as to why n or n-1 or 2n-1 is divisible by 3. I need to think about it more.