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A natural number N has 4 factors , Sum of the factors of N

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A natural number N has 4 factors , Sum of the factors of N  [#permalink]

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New post Updated on: 31 Jul 2013, 09:44
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A natural number N has 4 factors , Sum of the factors of N excluding N is 31, How many Values of N is possible

A)2
B)3
C)4
D)6
E)7

Originally posted by abhishekkhosla on 31 Jul 2013, 09:28.
Last edited by Zarrolou on 31 Jul 2013, 09:44, edited 2 times in total.
Edited the question.
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Re: A natural number N has 4 factors , Sum of the factors of N  [#permalink]

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New post 31 Jul 2013, 09:44
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abhishekkhosla wrote:
A natural number N has 4 factors , Sum of the factors of N excluding N is 31, How many Values of N is possible

A)2
B)3
C)4
D)6
E)7


A number has four factor is two cases: it's a prime number raised to the third power, so \(N=n^3\).
So in this case the factors are \(n^3\), \(n^2\), \(n\) and \(1\). Their sum is 31, so \(n^2+n+1=31\), \(n=\)5 and \(n^3=125\) or n=-6, but the natural numbers are positive so this is not valid. ( and 6 is not prime )

The other valid case is the one in which the number \(N=a*b\), where a and b are prime.
In this case the factors are \(1,a,b,a*b\), their sum is \(a+b+1=31\) or \(a+b=30\), where a and B are prime.
This sum is valid if the numbers \(a,b\) are \(7,23\) or \(13,17\) or \(11,19\). (please note that 1 and 29 is not a valid option)
So N can be \(7*23=161\) or \(13*17=221\) or \(11*19=209\).

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Re: A natural number N has 4 factors , Sum of the factors of N  [#permalink]

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New post 31 Jul 2013, 09:59
Kudos thanks for the explnation . I procedded this way

N= a.b or a^3 as you said a^3 is not possible so N=a.b
sum of factors=(a^2-1)\(a-1)*b^2-1\(b-1) = (a+1)(b+1)
Thus (a+1)(b+1)-ab=31 thus a+b=30 but after that i got confused but now i got it thanks
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Re: A natural number N has 4 factors , Sum of the factors of N   [#permalink] 31 Jul 2013, 09:59
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