Find the four digit number? Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.
HINT:

So, we have $$1000a+100a+10b+b=11(100a+b)$$

$\implies 100a+b$ must be divisible by $11\implies 11|(a+b)$ as $100\equiv1\pmod{99}$

As $0\le a,b\le 9, 0\le a+b\le 18\implies a+b=11$

$$\implies11(100a+b)=11(100a+11-a)=11^2(9a+1)$$

So, $9a+1$ must be perfect square
We know that the only square that has last two digits as same is $12^2=144$. For a four digit perfect square it should either be $(50-12)^2$ or $(50+12)^2$ or $(100-12)^2$  

$38^2= 1444 $
$62^2= 3844$
$88^2= 7744 $

As per the question we know $88^2$ is the perfect square that has first 2 digits &amp; last 2 digits as the same.