Numbers-5 (CAT-2007)

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 11² ie.,121

The other multiplicand should also be a perfect square.

Now we will try out some possibilities:

121 * 2² = 484 => option is out as its not a 4 digit number

121 * 3² = 1089 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 4² = 1936 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 5² = 3025 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 6² = 4356 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 7² = 5929 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 8² = 7744 => This suits

121 * 9² = 9801 => option is out as first and second digits are not same

as is the case for third and fourth digits also.

121 * 10² = 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)