Vtyjkyuk

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 $11implies 11|(a+b)$ as $100equiv1pmod99$

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

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

So, $9a+1$ must be perfect square

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!

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.

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.

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.

Given $overlineaabb=1100a+11b=k^2$, consider mod $4$:
$$kequiv 0,1,2,3 pmod4 \
k^2equiv 0,1 pmod4\
1100a+11bequiv 3bequiv 0,1 pmod4 Rightarrow b=0,3,4,7,8 (1)$$
Also, the last digit of $k^2$ can be:
$$b=0,1,4,5,6,9 (2)$$

Hence, from $(1)$ and $(2)$:
$$b=0 textor 4.$$
And:
$$k^2=1100a+11b=11(100a+b)=11^2cdot 9a+11(a+b) Rightarrow a+bequiv 0 pmod11.$$
So, $overlineaabb=7744$.

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.