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Problem of the Day 30.07.09

with 7 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 Implex

July 30, 2009 at 1:03 pm

7 Responses

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  1. N=126; (1)


    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)


    August 4, 2009 at 7:10 pm

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


    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..


    August 20, 2009 at 7:50 pm

  5. Let the least number be N, 1 is its least divisor.Let 2nd,3rd and 4th least divisors be x,y and z respectively. We consider the following values of divisor a and the corresponding values of a^3, from x,y and z exactly 1 or all 3 are odd.(N is even) a=1:a^3=1 a=2:a^3=8 a=3:a^3=27 a=4:a^3=64 a=5:a^3=125 a=6:a^3=216 For x, y and z=(2,3,4),2*N=100(i.e.N=50). But 3 is not a divisor of 50. For x,y,z=(2,3,6),2*N=252(i.e.N=126) and the 1,2,3,6 are four least distinct divisor of 126.The required number is 126. The sum of digits is 9.


    August 15, 2013 at 5:01 pm

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