Advent calendar 2025

For a multiplication table from 1 to \(n\), the sum of all the numbers in the top row is \(1+...+n\), the sum of all the numbers in the second row is \(2(1+...+n)\), the sum of all the numbers in the third row is \(3(1+...+n)\), ..., and the sum of all the numbers in the \(n\)th row is \(n(1+...+n)\). The sum of all the numbers is therefore \((1+...+n)(1+...+n)=n^2(n+1)^2/4\) and the sum of the numbers in the bottom row is \(n^2(n+1)/2\).

If the sum of all the numbers is 2025, then \(n=9\) and so the sum of the numbers in the bottom row if 405.