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

As printed, the claim is false: $70$ is 7-cyclic-divisible because its rotations are $70$ and $07=7$, but its digits are not all equal to $7$ and its length is not a multiple of $6$.

The standard rotation argument proves the likely intended statement. If an $L$-digit integer and all its rotations are divisible by $7$, then for each leading digit $a$,is divisible by $7$. The multiplicative order of $10$ modulo $7$ is $6$. If $6\nmid L$, then $7\nmid 10^L-1$, forcing every digit $a$ to be divisible by $7$, hence to be either $0$ or $7$. Thus the valid conclusion is:The counterexample shows why “all its digits are equal to 7” cannot replace the first alternative.

Solved by gpt-5.6-sol high.