Why does every number of shape ababab is divisible by $13$?
Solution 1:
Note that $$ [ababab] = a\times 101010 + b \times 10101 = 13 (7770a + 777b) $$
Also noteworthy: $$ 10101 = 1 + 10^2 + 10^4 \equiv \\ 1 + 3^2 + 3^4 = 1 + 9 + 9^2 \equiv\\ 1 +(-4) + (-4)^2 = 1 - 4 + 16 = 13 \equiv 0 $$ where $\equiv$ indicates equivalence modulo $13$.
Solution 2:
These numbers are of the form $(10a+b)\cdot 10101$, and $10101=13\cdot 777$.
Solution 3:
Note: $$ab=a\times 10+b\times 1$$
so $$ab\times 100=ab00$$ therefore
$$abab=ab00+ab=ab(100+1)=ab\times 101$$ $$abab00=ab\times101\times100=ab\times10100$$ $$ababab=ab\times10101=ab\times3\times7\times13\times37$$
Hence it is divisible by $3$,$7,13$ and $37$.
Solution 4:
A number $ABCDEF$ is divisible by $13$ if and only if $ABC-DEF$ is divisible by $13$.
Note that $\small{ABA-BAB=100(A-B)+10(B-A)+1(A-B)=91A-91B=13(7A-7B)}$.
Therefore $ABA-BAB$ is divisible by $13$.
Therefore $ABABAB$ is divisible by $13$.
Solution 5:
And not only that: Each such number is also divisible by the other primes 3, 7 and 37. Just factor the number 10101! Your specimen number is any 2-digit number $\times 10101\,$.