= 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 $a$,
$$
10c_j(n)-c_{j+1}(n)=a(10^L-1)
$$
is divisible by $7$. The <multiplicative order> of $10$ modulo $7$ is $6$. If $6\nmid L$, then $7\nmid10^L-1$, forcing every digit $a$ to be divisible by $7$, hence to be either $0$ or $7$. Thus the valid conclusion is:
$$
\boxed{\text{either every digit is }0\text{ or }7,
\quad\text{or }6\mid L.}
$$
The counterexample shows why “all its digits are equal to 7” cannot replace the first alternative.
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