## Problem Of The Day 04.07.2009

Let *aabb* be a 4-digit number (a≠0). How many such numbers are perfect squares?

A) 0 B) 1 C) 2 D) 3 E) 4

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the answer is option (B) 1

coz..d number shud in d form of 11x^2…

aa=11..

den bb must be in d form a0b=704

…

simply (88)^2=7744

anupJuly 3, 2009 at 6:32 pm

yeah right

it came in CAT 2007

RahulJuly 3, 2009 at 6:38 pm

There is a neat trick to it.

(25±x)², (50±x)², (100±x)² have last two digits of the same form.

now only the square of 12 has last two digits which are same

hence, the squares of 12, 38, 62 and 88 have 44 as last two digits.

only 88 is a multiple of 11.

Hence 1 solution

RahulJuly 3, 2009 at 9:05 pm

thnx rahul

anupJuly 4, 2009 at 6:10 am