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abhishekkhosla
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A natural number N has 4 factors , Sum of the factors of N Updated on: 31 Jul 2013, 09:44
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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85% (hard)

Question Stats:

27% (01:55) correct 73% (02:22) wrong based on 33 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
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C
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Zarrolou
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A natural number N has 4 factors , Sum of the factors of N 31 Jul 2013, 09:44
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abhishekkhosla
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
Show more

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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A natural number N has 4 factors , Sum of the factors of N 31 Jul 2013, 09:59
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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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A natural number N has 4 factors , Sum of the factors of N 14 Oct 2020, 02:09
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abhishekkhosla
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
Show more

Should the answer not be "3" values: 161, 221 and 209? What is the 4th value of N?

So N can be \(7*23=161\) or \(13*17=221\) or \(11*19=209\)
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A natural number N has 4 factors , Sum of the factors of N 14 Oct 2020, 04:39
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PANKAJ0901
Zarrolou
abhishekkhosla
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

Should the answer not be "3" values: 161, 221 and 209? What is the 4th value of N?

So N can be \(7*23=161\) or \(13*17=221\) or \(11*19=209\)
Show more
4th value is 125
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A natural number N has 4 factors , Sum of the factors of N 14 Oct 2020, 08:17
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This isn't really in the style of an authentic GMAT question, but I'm not sure why the 'out of scope - too hard' tag has been applied to it. If a number has four factors, its prime factorization must either look like p^3 or like pq, where p and q are different primes.

If a number's prime factorization is p^3, then its factors are 1, p, p^2 and p^3. Ignoring the number itself (p^3), the sum of its factors is 1 + p + p^2, and if that equals 31, then p + p^2 = 30. Only one prime could possibly satisfy that equation (since the left side gets larger the larger p is, when p is positive) and we can quickly see p = 5 works, so that's the only solution in this case.

If a number's prime factorization is pq, then its factors are 1, p, q and pq. Ignoring the number itself (pq), the sum of its factors is 1 + p + q, and if that equals 31, then p + q = 30. So we'll also have solutions for any pair of distinct primes that sum to 30. So our number could have any of the prime factorizations (13)(17), (11)(19) or (7)(23), which all must be different numbers by the fundamental theorem of arithmetic.

So the answer is four.
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