Question 22

# Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

Solution

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $$(50 \pm a)$$ and $$100 - a$$ where $$0 \leq a \leq 100$$
So, the square would be of form $$(50 \pm a)^2 = 2500 + a^2 \pm 100a$$ or $$(100 - a)^2$$ i.e. $$10000 + a^2 + 200 a$$

In both cases, only $$a^2$$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

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