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

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Joined: 02 Jun 2013
Posts: 18
A natural number N has 4 factors , Sum of the factors of N  [#permalink]

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Updated on: 31 Jul 2013, 09:44
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45% (medium)

Question Stats:

29% (00:49) correct 71% (02:34) wrong based on 8 sessions

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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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31 Jul 2013, 09:44
1
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 $$11*19=209$$.

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

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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
Re: A natural number N has 4 factors , Sum of the factors of N   [#permalink] 31 Jul 2013, 09:59
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