<s>[INST] Someone who is calculating $43434343^2$. The answer is $18865ab151841649$, where the two digits $a$ and $b$ were lost. So I used congruence $\bmod 9$ and $\bmod 11$. First, $43434343^2 \pmod{11}$ is $5$ and I did the answer $18865ab15184169 \pmod{11}$ and got $-b+a+3 \pmod{11}$. I also did $\bmod 9$ and got $a+b+67 \pmod 9$. I am confused as to what to do next.
 [/INST] The simplest is probably to look modulo $99$.
Since $100 \equiv 1 \pmod {99}$, 
$43434343^2 \equiv (43+43+43+43)^2 \equiv 172^2 \equiv 73^2 \equiv 5329 \equiv 82 \pmod {99} $
$18865ab151841649 \equiv 18+86+(50+a)+(10 b+1)+51+84+16+49 \equiv 355+ba \equiv 58+(10b+a) \pmod {99}$.
Hence $(10b+a) \equiv 24 \pmod {99}$. Luckily, we didn't get zero, so there is only one possibility : $b=2$ and $a=4$.
</s>