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Problem Of The Day 04.07.2009

with 4 comments


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

Written by Implex

July 3, 2009 at 4:55 pm

4 Responses

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  1. 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

    anup

    July 3, 2009 at 6:32 pm

  2. yeah right
    it came in CAT 2007

    Rahul

    July 3, 2009 at 6:38 pm

  3. 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

    Rahul

    July 3, 2009 at 9:05 pm


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