$a,b,c \in$ {0,1,2,...,9} with at least one of $a,b,c$ nonzero. Prove that the six-digit integer $abcabc$ is divisible by at least 3 distinct primes.
My thinking is not to use induction as there is a definite size to the integer.
So $abcabc$ comes to $c1001 + b10010 + a100100$ with at least one of them nonzero. This factors to $7^{1}11^{1}13^{1}c + 7^{1}11^{1}13^{1}2^{1}5^{1}b + 7^{1}11^{1}13^{1}2^{2}5^{2}a$ which factors to $7^{1}11^{1}13^{1}(c + 2^{1}5^{1}b + 2^{2}5^{2}a$) so with one of $a,b,c$ nonzero $abcabc$ is divisible by at least 3 primes $7,11,13$.
$$\overline{abcabc}=\overline{abc}(1001)$$ as: $$\overline{abcabc}\\=10^5a+10^4b+10^3c+10^2a+10^1b+10^0c\\=10^2a(10^3+1)+10^1b(10^3+1)+10^0c(10^3+1)\\=(10^2a+10b+c)(10^3+1)$$ Since $1001$ is divisible by $7,11$ and $13$; you're done.