Divisibility tests, finding the value of two missing numbers in a message.

Question

I was checking the following divisibility tests exercise:

You receive a message from an extraterrestrial alien, who is calculating $43434343^{2}$. The answer is $18865ab151841649$, where the two digits represented as a and b were lost in transmission. Use congruences $mod$ $9$ and $mod$ $11$ to determine the answer to this fundamental problem.

I've been trying to start changing the required numbers on $18865ab151841649$, the sum term by term is $67$, so we need

$a+b$ $=$ $5$ to reach $72$ multiple of $9$, we have six couples:

$(3,2)$ $(2,3)$ $(4,1)$ $(1,4)$ $(5,0)$ $(0,5)$

After that what I tried to do is to check each couple with the divisibility test for 11, but none of them worked by substracting the amount on even positions to odd positions. Any help will be really appreciated.

Answer 1

You need $a+b\equiv 5 \bmod 9$, not just $a+b=5$, so $a+b=14$ would be an option too.

Answer 2

You need to consider congruence for both $43434343^{2}$ and $18865ab151841649$.

Consider the first number: $$\begin{align}43434343^{2}&\equiv 28^2\equiv 1^2\equiv 1 \pmod{9};\\ 43434343^{2}&\equiv (43\cdot 73\cdot 101\cdot 137)^2\equiv ((-1)\cdot (-4)\cdot 2\cdot 5)^2\equiv (4\cdot (-1))^2\equiv 5 \pmod{11}.\end{align}$$ Now consider the second number: $$\begin{align}18865ab151841649&\equiv a+b+67\equiv a+b+4\equiv 1\pmod{9} \Rightarrow a+b\equiv 6\pmod{9} \Rightarrow \\ (a,b)&=(9,6),(8,7),(7,8),(6,9),(6,0),(5,1),(4,2),(3,3),(2,4),(1,5),(0,6);\\ 18865ab151841649&\equiv \pm[(a+35)-(b+32)]\equiv \pm[a-b+3]\equiv 5\pmod{11} \Rightarrow \\ 1) \ a-b&\equiv 2\pmod{11} \ \ \text{OR} \ \ 2) \ b-a\equiv 8\pmod{11} \Rightarrow \\ 1) \ (a,b)&=(4,2) \ \ \text{OR} \ \ 2) \ (a,b)\in \emptyset. \end{align}$$ So, the final answer is: $(a,b)=(4,2)$.

Indeed: $$43434343^{2}=18865\color{red}{42}151841649.$$