Why does every number of shape ababab is divisible by $13$? Why does it seems like every number $ababab$, where $a$ and $b$ are integers $[0, 9]$ is divisible by $13$? 
Ex: $747474$, $101010$, $777777$, $989898$, etc...
 A: 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 $37$.
A: 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 $13$.
A: $abcabc=abc(10^3+10^6)=abc\cdot 1001000=abc\cdot13\cdot77000$.
REMARK.-It is easy generalisable to $abcabcabcabcabc.....abcabc$ for $2n$ times abc.
A: 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 $\times 10101\,$.
A: 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 $13$.
A: $ababab$ is a multiple of $10101$, which in turn is a multiple of $13$.
A: Hint $\,\ {\rm mod}\ 13\!:\,\ 10\equiv 6^2\,\Rightarrow\, 10^6\equiv 6^{12}\equiv 1\,$ by little Fermat.
Hence $\, 0\equiv 10^6-1 \equiv (10^2\!-1)(10^4\!+10^2\!+1) \equiv 99\cdot 10101$
Thus $\,99\not\equiv 0\,\Rightarrow\,10101\equiv 0\,\Rightarrow\, ab\cdot 10101 = ababab\equiv 0,\,$ for $\, 0 \le ab \le 99$
A: These numbers are of the form $(10a+b)\cdot 10101$, and $10101=13\cdot 777$.
A: $100^0 \equiv 1 \bmod 13$
$100^1 \equiv 9 \bmod 13$
$100^2 \equiv 3 \bmod 13$
$(ababab)_{10}=c100^2+c100+c =c(100^2+100+1) \equiv c(3+9+1) \equiv 0 \bmod 13$, 
where $c=10a+b$.
