$abc0a$c is a six digit perfect square number which is divisible by 5 and 11. Find out the number.
Source: Bangladesh Math Olympiad 2016 Junior Category
I tried but failed to find any relation between 5, 11 and a square number. I have also seen a similar question with 13 and 15. I found that last digits of the square number are 0 or 5. Can anyone give me necessary hints to solve this problem.
On
Since it's a perfect square divisible by $5$ and $11$, it is divisible by $25$ and $121$. This immediately tells you that the last two digits $ac$ is one of $00$, $25$, $50$ or $75$.
Using divisibility test for $11$:
Case $00$: $0b0000$ is not divisible by $11$ for any $b$.
Case $25$: $2b5025$ gives you $b = 4$ if it is divisible by $11$ and is a perfect square.
Case $50$: $5b0050$ is not divisible by $11$ for any $b$.
Case $75$: $7b5075$ gives you $b = 3$ if it is divisible by $11$ but it is not a perfect square.
So the answer is 245025.
We need to use divisibility rules.
First, because the number is divisible by 5, we know that $c=0,5$.
But, if $c=0$, then $a=0$, since that implies that the 6 digit number is divisible by 10, implying that it also divides 100. But, this would make the number not a 6 digit number. Hence, $c=5$.
But, if $c=5$, $a=2$, since that means that the number divides 5 but not 2, implying that the last two digits of the number must be 25 (any odd number squared is congruent to 1 mod 4). So, we know our number is of the form $2b5025$.
Next, since the number is divisible by 11, $a+c+a-b-c=2a-b$ is divisible by 11. So, 11 divides $4-b$, which implies $b=4$. So, our final answer is $$245025$$