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Posts: 15
Updated on: 31 Jul 2013, 08:44
Originally posted by abhishekkhosla on 31 Jul 2013, 08:28.
Last edited by Zarrolou on 31 Jul 2013, 08:44, edited 2 times in total.
Last edited by Zarrolou on 31 Jul 2013, 08:44, edited 2 times in total.
Edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
26% (01:55) correct
74%
(02:17)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
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)2
B)3
C)4
D)6
E)7
abhishekkhosla
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
Posts: 15
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
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
Pankaj0901
Joined: 18 Dec 2018
Last visit: 17 Dec 2022
Posts: 399
Location: India
WE:Account Management (Hospitality and Tourism)
Posts: 399
Zarrolou
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 \(11*19=209\)
Posts: 55
PANKAJ0901
4th value is 125
IanStewart
Expert
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 10 Sep 2026
Posts: 4,146
Location: United States
GMAT 1: 780 Q51 V47
Expert reply
GMAT 1: 780 Q51 V47
Posts: 4,146
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.
_________________
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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