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Find out a 4 digit perfect square number in which the first two digits are same. Again the last two digits are also same.

Jacob Frye
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    The solution is $88^2=7744$. Note that the square must be divisible by $11$, hence you only have to check the numbers $11^2,22^2,33^2,\cdots ,99^2$ – Peter Apr 11 '18 at 09:58

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$$x^2 = \overline{aabb}$$

$$x^2 = 1000 a + 100a + 10b + b = 1100a + 11b = 11(100a + b)$$

$$x^2 = 11 \cdot \overline{a0b}$$

11 divides $x^2$ so it divides $x$ as well. It also means that 11 divides $\overline{a0b}$. Divisibility rule says that 11 also divides $a+b$.

You don't have to check all combinations because a perfect square ends only with 1, 4, 5, 6 and 9. The first option ($b=1$) can be discarded because $a<10$.

It means that you have to test only 4 possible solutions: 7744, 6655, 5566 and 2299. Only 7744 is a valid solution.

Saša
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