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.
 A: If we let the four-digit number be XXYY, then this number can be expressed as:
$$1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2$$ 
(since it's a perfect square)
In order for this to be true, $100X + Y$ must be the product of $11$ and a perfect square, and looks like $X0Y$. So now our question is "which product of $11$ and a perfect square looks like $X0Y$?" We can test them:
$$11 \cdot 16 = 176\\
 11 \cdot 25 = 275\\
 11 \cdot 36 = 396\\
 11 \cdot 49 = 593\\
 11 \cdot 64 = 704\\
 11 \cdot 81 = 891$$
The only one that fits the bill is $704$. This means there is only one four-digit number that works, and it's $7744$. Enjoy!
A: I recommend programming when numbers are so low.
Here is a Python solution:
>>> list(filter(lambda n: str(n**2)[0] == str(n**2)[1] and \
                          str(n**2)[-1] == str(n**2)[-2], 
                range(int(1000**0.5),int(10000**0.5))
               )
         )
[88]
>>> 88**2
7744

Note that I broke the line for easier readability.
So 7744 is the only solution.
A: 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
A: Given $\overline{aabb}=1100a+11b=k^2$, consider mod $4$:
$$k\equiv 0,1,2,3 \pmod{4} \\
k^2\equiv 0,1 \pmod{4}\\
1100a+11b\equiv 3b\equiv 0,1 \pmod{4} \Rightarrow b=0,3,4,7,8 \ \ \ \ \ \ (1)$$
Also, the last digit of $k^2$ can be:
$$b=0,1,4,5,6,9 \ \ \ \ \ \ (2)$$
Hence, from $(1)$ and $(2)$:
$$b=0 \ \ \text{or} \ \ 4.$$
And:
$$k^2=1100a+11b=11(100a+b)=11^2\cdot 9a+11(a+b) \Rightarrow a+b\equiv 0 \pmod{11}.$$
So, $\overline{aabb}=7744$.
A: you'd have to just analyze the digits of some general square of a two digit number: 
(10x + y)² = 100x² + 2xy + y² 
It turns out, x=y=8.
A: The Number is of the form 1000A + 100A + 10B + B = 11( 100A + B ) = 11 ( 99A + A + B )
Since it is a perfect square number , (99A + A + B) should be divisible by 11 hence (A + B) is divisible by 11....(i)
Any perfect square has either the digits 1 , 4 , 9 , 6 , 5 , 0 at the units' place , ...(ii)
Only numbers which satisfy property i and ii are enlisted:
7744
2299
5566
6655
[Note that (A + B) being divisible by 11 was a crucial property to note]
Clearly , Only 7744 is a perfect square number.
A: 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 & last 2 digits as the same.
