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Problem of the Day 30.07.09

with 6 comments

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

Written by Rahul

July 30, 2009 at 1:03 pm

6 Responses

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  1. N=126; (1)

    anonymous

    August 1, 2009 at 5:16 pm

    • can u plz explain the method ?

      nikhil jangi

      August 3, 2009 at 11:51 am

  2. 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)

    anonymous

    August 4, 2009 at 7:10 pm

  3. hit N trial method N u will get 126.

    ANKIT PANGHAL

    August 10, 2009 at 10:09 am

    • 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.

      A nkur Dudeja

      August 18, 2009 at 1:56 pm

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

    Prateek

    August 20, 2009 at 7:50 pm


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