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?

(1) 3

(2) 2

(3) 4

(4) 0

(5) 1

Solution follows here:**Solution:**

Let the number be xxyy.

xxyy = 1000x+100x+10y+y = 1100x+11y = 11(100x+y)

**So the number is multiple of 11. To be a perfect square, it should be a multiple of 11**

^{2}ie.,121**The other multiplicand should also be a perfect square.**

Now we will try out some possibilities:

121 * 2

^{2}= 484 => option is out as its not a 4 digit number
121 * 3

^{2}= 1089 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 4

^{2}= 1936 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 5

^{2}= 3025 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 6

^{2}= 4356 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 7

^{2}= 5929 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 8

^{2}= 7744 =>**This suits**
121 * 9

^{2}= 9801 => option is out as first and second digits are not same
as is the case for third and fourth digits also.

121 * 10

^{2}= 12100 => From here onwards the numbers are out of range
(five digit numbers started)

The only four digit number suiting the requirement is ‘7744’

**Answer (5)**

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