# 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.

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Am I right to say that b cannot equal 1? –  Adam Nov 26 '13 at 10:12

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

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@user93470, we can test for $0\le a\le 9\implies a=0,7$ –  lab bhattacharjee Nov 18 '13 at 8:19
I am looking for something more concrete... –  user93470 Nov 18 '13 at 8:32
–  lab bhattacharjee Nov 18 '13 at 8:38

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.

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