An unavoidable case analysis

Archiving old manuscripts, I found at the end of EWD766 "An educational stupidity" of more than 20 years ago the following exercise:

"Prove that none of the decimal numbers 1001, 1001001, 1001001001, 1001001001001, ... is prime." [It is not clear why I underlined "none". EWD]

Here is a proof. Denoting by ^k*<string>^* the concatenation of k copies of the digit string enclosed, we deal with the decimal numbers

^k*001^* for k ≥ 2

If k mod 3 = 0, the number is, according to the traditional 3-test, divisible by 3 because the sum of its decimal digits, which equals k, is divisible by 3.
If k mod 3 ≠ 0, we see by generalising the traditional 9-test to the ^k*9^*-test, that the number reduced modulo ^k*9^* equals ^k*1^*. Hence the number is divisible by ^k*1^* (because ^k*9^* is). (Note that this observation does not exclude primality for k=1.)

In argument (i), the crux is the validity of the 3-test, which is valid whenever base mod 3 = 1, a condition that base 10 (= ten) satisfies. The conclusion has nothing to do with the accident that the length of the repeated string happens to be 3: for k mod 3 = 0, also ^k*00001^* is divisible by 3.

In argument (ii) we generalized the 9-test because the problem was about decimal numbers, but for any base B (B ≥ 2) there is a (B−1)-test, and ^k*1^* divides ^k*(B−1)^*. The conclusion that the number reduced modulo ^k*(B−1)^* yields ^k*1^*, however, depends on the fact that k has no factor in common with the length of the iterated string. Argument (ii) is base independent: interpreting ^*k001^* and ^*k1^* in binary, we find for instance that for k=4 and k=5, 585 is divisible by 15 and 4681 by 31.

Argument (i) relies on a relation between k and the base of the number system, (ii) on a relation between k and the length of the iterated string.