$5$ and $11$ divides a perfect square $abc0ac$. What is the number?

Question

$5$ and $11$ divides a perfect square $abc0ac$. What is the number?

I started this way - expressing the number as $(10^5+10)a+(10^4)b + (10^3+1)c$
Which fooled me. How can I start?
P.S: This is a problem from BdMO-2016 regionals.

Answer 1

Because $25$ divides N it follows $ac \in \{ 00, 25, 50, 75\}$. Because $11$ divides N it follows that the alternating sum of the digits in the number is divisible by 11, therefore $11 | 2a -b$.

Now just take each possibility for $ac$

Answer 2

I will post an answer for completion.
If the number is divisible by $5$ then $c=5$ or $c=0$. It cannot equal $0$ by Huang's comment. Therefore $c=5$ and because a perfect square ending in $5$ necessarily ends in $25$ we know that $a=2$.
If a number is divisible by $11$ then the alternating sum of its digits: $a-b+c-0+a-c=2a-b=4-b=11n$ (i.e. the sum must be a multiple of $11$). Because $0\le b \le 9$, then $n$ can only equal $0$, from which we know that $b=4$ and the number equals $$245025=495^2$$