Simple but fun
Let \(n\) be a positive odd integer and \(a_1,\ldots,a_n\) a permutation of \(1,\ldots,n\). Prove that
\[\prod_{i=1}^{n}(a_i-i)\]is even.
Hint
Could all \(n\) differences be odd?
Solution
Permuting the numbers preserves their sum, so
\[\sum_{i=1}^{n}(a_i-i)=0.\]If every difference were odd, their sum would be odd because \(n\) is odd. Thus at least one factor is even, and so is the product.