Puzzles

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.

The product of the numbers in the red boxes is 336.

Every number except the powers of 2 can be written as the sum of two or more consecutive positive integers. The smallest three-digit power of 2 is 128.

Why can every number that isn't a power of 2 be written as the sum of two or more consecutive positive integers?

The number of ways to form sets like this containing the numbers 1 to \(n\) is the \((n+2)\)th Fibonacci number. Therefore there are 987 ways for the numbers from 1 to 14.

Can you show why the solution is the Fibonacci numbers in a similar way to in the solution to the puzzle on 6 December?

8×7×5×3=840. This is also a multiple of 4, 6, 2, and 1. If anything is removed from the product it will no longer be a multiple of all the numbers.

414