## 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**

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N=126; (1)

anonymousAugust 1, 2009 at 5:16 pm

can u plz explain the method ?

nikhil jangiAugust 3, 2009 at 11:51 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)

anonymousAugust 4, 2009 at 7:10 pm

hit N trial method N u will get 126.

ANKIT PANGHALAugust 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 DudejaAugust 18, 2009 at 1:56 pm

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

PrateekAugust 20, 2009 at 7:50 pm

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.

ankeshgupta09August 15, 2013 at 5:01 pm