Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

3 Answers 3

up vote 9 down vote accepted

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

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

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.

share|improve this answer

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!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.