Thursday, July 16, 2009

Problem of the day 16.7.09

What is the sum of the digits of a two digit number which is 32 less than the square of the product of its digits?

A. 12                   B. 11                 C. 10 .                     D. 9                           E.      8
Posted by Unknown at 10:58 AM
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10 comments:

  1. anonymousJuly 16, 2009 at 1:57 PM

    8

    ReplyDelete
  2. HariJuly 16, 2009 at 2:09 PM

    I got the same by trial & error.. trying dif combs till i got the answer as 17 .
    Could you please elaborate the method that you might have used ?

    ReplyDelete
  3. RahulJuly 16, 2009 at 2:18 PM

    I started with 36-32=4 not works
    49-32=17 which is 7^2
    so we r done..

    it can be done algebraically, but its a bit tedious

    ReplyDelete
  4. HariJuly 16, 2009 at 2:36 PM

    ok thanks man ...

    ReplyDelete
  5. nikhilJuly 16, 2009 at 5:20 PM

    answer : 17

    Method : tens place cannot be 2 since after 24 the differences hugely . so tens place has to be one and becomes quite apparent that it has among 15,16,17, 18 .

    ReplyDelete
  6. RahulJuly 16, 2009 at 7:08 PM

    Nice, Nikhil

    ReplyDelete
  7. kuldeepJuly 20, 2009 at 6:41 PM

    8

    ReplyDelete
  8. shivamJuly 24, 2009 at 2:00 PM

    method
    max possible number 99+32=131 and minimum is 10+32=42
    we have to watch for squares b/w them.
    we are left with 36/49/64/81

    (1x7)^2=49 =>49-32=17

    ReplyDelete
  9. rahulJuly 25, 2009 at 8:21 PM

    hai shivam , yours approach is excellent

    ReplyDelete
  10. AakashAugust 11, 2009 at 4:46 PM

    i started out wit 12 but due to huge diff. shifted directly 2 option e and gotten the ans by trial n error... bt i suppose its a bit stupid coz its tym consuming... it ws nice anyways.

    ReplyDelete

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