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.
 A: 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 $a=4$.
A: $\let\cong\equiv$
So you know that
$$43434343^2 \cong 5 \pmod{11}$$
and
$$43434343^2 \cong 1 \pmod{9}$$
and you found that
$$\begin{aligned}
18865ab15184169 &\cong a-b+3 \pmod{11}, \\
18865ab15184169 &\cong a+b+67 \pmod{9}.
\end{aligned}$$
You want $18865ab15184169$ to be equal to $43434343^2$, so the congruence classes of these two numbers in both $(\bmod 9)$ and $(\bmod 11)$ also need to match:
$$\begin{aligned}
a-b+3 &\cong 5 \pmod{11}, \\
a+b+67 &\cong 1 \pmod{9}.
\end{aligned}$$
This can be simplified further as
$$\begin{aligned}
a-b &\cong 2 \pmod{11}, \\
a+b &\cong 6 \pmod{9}.
\end{aligned}$$
At the same time, both $a$ and $b$ are just decimal digits and thus constrained to $\{0,1,2,3,4,5,6,7,8,9\}$. This leaves not many possibilities how to satisfy the first congruence: either $a-b=2$ (note that this is an actual equality, not a congruence) or $a-b=-9$, which could only mean $a=0$, $b=9$.
We discard the latter quickly by finding that $0+9 \not\cong 6 \pmod{9}$. The other possibility would mean $a = b+2$, whatever $b$ is, so plug that into the remaining congruence:
$$\begin{aligned}
(b+2)+b &\cong 6 \pmod{9} \\
2b + 2 &\cong 6 \pmod{9} \\
2b &\cong 4 \pmod{9}
\end{aligned}$$
Out of all digits, only $b=2$ can satisfy this. Then use $a = b+2$ again to find $a$ and you're done.
Another way:
When at the point where you have the congruences
$$\begin{aligned}
a-b &\cong 2 \pmod{11}, \\
a+b &\cong 6 \pmod{9},
\end{aligned}$$
see what we can find for the sum of the two LHS's and for their difference:
$$\begin{aligned}
2a &\cong 2+6 \pmod{11}, \\
2b &\cong 6-2 \pmod{9}.
\end{aligned}$$
The solution $a=4$, $b=2$ follows immediately.
