Thursday, July 30, 2009

Problem of the Day 30.07.09

Find the sum of the digits of the least natural number N, such that the sum of the cubes of the four smallest distinct divisors of N is 2N?



1)  9                                2) 8                             3) 7                    4) 6                      5) 10

Posted by Unknown at 6:33 PM
Labels: , , , , , , , , , ,

6 comments:

  1. anonymousAugust 1, 2009 at 10:46 PM

    N=126; (1)

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  2. nikhil jangiAugust 3, 2009 at 5:21 PM

    can u plz explain the method ?

    ReplyDelete
  3. anonymousAugust 5, 2009 at 12:40 AM

    I used hit and trial, guessed it would be the quickest approach to solve it.
    Start with 1,2,3,4 then 1,2,3,6 (which fetches the answer)

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  4. ANKIT PANGHALAugust 10, 2009 at 3:39 PM

    hit N trial method N u will get 126.

    ReplyDelete
  5. A nkur DudejaAugust 18, 2009 at 7:26 PM

    If the number is even then 1 and 2 are gonna be its two divisors.The remaining 2 divisors could be p and 2p for sure then(taking p as a prime number).We can check p for 3,5,7 and 3 will give the answer by meeting all the conditions.

    ReplyDelete
  6. PrateekAugust 21, 2009 at 1:20 AM

    Rahul pls put the approach to this question and answer to it..

    ReplyDelete

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