Find the number $ccbb$

Question

If the number has $4$ digit $ccbb$ and it's full square, then find that number.

I have tried and I got $88^2=7744$ but my way has no prove for it, if any one have away, I'll appreciate it.

Answer 1

$ccbb$ (in decimal notation) is $1100 c + 11 b = 11(100 c + b)$. If this is a square $x^2$, then $x$ must be divisible by $11$, so $x = 11 a$, and then $11 a^2 = x^2/11 = 100 c + b$. Now $100 \equiv 1 \mod 11$, so $0 \equiv 100 c + b \equiv c + b \mod 11$.

Now $1 \le c + b \le 18$ so we must actually have $c + b = 11$, and then $100 c + b = c + b + 99 c = 11 + 99 c = 11(1 + 9 c)$. Thus $1 + 9 c$ is a square.

The possible values of $c$ are $2,3, \ldots, 9$ corresponding to $1+9c = 19, 28, 37, 46, 55, 64, 73, 82$, and the only one of those that is a square is $64 = 8^2$. So we must have $c = 7$, and then $b=4$.

Answer 2

Note that $ccbb=1000c+100c+10b+b=1100c+11b=11(100c+b)$. Therefore, $11\mid(100c+b)$.

In other words $100c+b\equiv 0\pmod{11}$. But $100\equiv 1\pmod{11}$. This implies $c+b\equiv 0\pmod{11}$. Notice that since $c$ and $d$ are both natural numbers, we must (?) have $c+b=11$. By trial and error you will get $c=7$ and $b=4$.

Answer 3

Since the number $ccbb = 1000c+100c+10b+b = 1100c+11b = 11(100c+b)$ is a multiple of $11$, and is a perfect square, its square root must also be a multiple of $11$.

Since $22^2 < 1000$ and $110^2 > 10000$, we only need to check the numbers $33^2 = 1089$, $44^2 = 1936$, $55^2 = 3025$, $66^2 = 4356$, $77^2 = 5929$, $88^2 = 7744$, and $99^2 = 9801$. There are only 7 of them to check, so this isn't hard.