## Thursday, 27 October 2011

### 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’