Divisibility by $2$ or $5$ depends only on the final digit. As the digits are cyclically rotated, every digit appears in the final position. Thus every rotation is divisible by $d\in\{2,5\}$ exactly when every digit is divisible by $d$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 5E / ii / Solution

 Solution (source code)