# 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
@labbhattacharjee, I didn't understood how did you did it. Please explain clearly! – Sagnik Saha Jun 27 '14 at 10:14
7744 is the answer that's sure! – Sagnik Saha Jun 27 '14 at 10:16

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!

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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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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.

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Can you be more specific on how to analyze the digits and conclude that $x=y=8$? Is that the only solution (and if yes, why)? I can't tell from your answer if you have a clever argument or if you just did a brute force search. (Also, you should replace 2 with 20 in your formula.) – Joonas Ilmavirta Sep 10 '14 at 19:14

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.

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