Senior Manager
Joined: 08 Jan 2018
Posts: 297
Location: India
Concentration: Operations, General Management
GMAT 1: 640 Q48 V27
GMAT 2: 730 Q51 V38
GPA: 3.9
WE:Project Management (Manufacturing)
Re: HOT Competition: If n is a positive integer, then n is a multiple of
[#permalink]
25 Aug 2020, 09:53
IMO A
If n is a positive integer, then n is a multiple of how many positive integers?
Note: n is a multiple of how many positive integers = No. of factor of n
If n = p^a x q^b x r^c, where p,q,r are prime numbers & a,b,c integers, then No. of factors of n= (a+1)(b+1)(c+1)
(1) n^5∗n^(−2) has exactly 5 divisors between 1 and itself, both not inclusive.
n^5∗n^(−2) = n^3 has 7 factors in total.
no. of factors = (a+1)(b+1)(c+1) = 7 = 7x1x1 => a=6 , b=c=0
So, n^3 must be of the form p^6 i.e n^3 = p^6 => n = p^2 (where p = prime number)
Since, n=p^2, so No. of factors of n=(2+1) =3 [Including 1 & no. itself]
Sufficient
(2) 3n^2 has exactly 8 divisors between 1 and itself, both not inclusive.
3n^2 has 10 factors in total.
Therefore, (a+1)(b+1)= 10= 10x1 = 2x5
Case 1: a=9 , b=0
In this case 3n^2 is of form p^9 , This is possible when n=3^4 , then 3n^2 = 3 x (3^4)^2 = 3^9
Therefore, n=3^4 has (4+1)= 5 Factors
Case 2: a=1 , b=4
In this case 3n^2 is of form p x q^4 , This is possible when p=3 & q=(Any prime no. except 3)^2
Here, n^2 = q^4 => n = q^2 (q = prime no. other than 3)
No. of factors of n = (2+1) = 3
Two possible values possible, therefore Insufficient
Posted from my mobile device