HOT Competition: If n is a positive integer, then n is a multiple of

Question

If n is a positive integer, then n is a multiple of how many positive integers?

(1)  n^5*n^{(-2)}  has exactly 5 divisors between 1 and itself, both not inclusive.

(2)  3n^2  has exactly 8 divisors between 1 and itself, both not inclusive.

GMATinsight · Accepted Answer

Answer: Option A Please check the video for the step-by-step solution. GMATinsight's Solution https://www.youtube.com/watch?v=vt21gnNY-HM Screenshot 2020-08-26 at 9.16.25 PM.png

NiftyNiffler · Answer

If n is a positive integer, then n is a multiple of how many positive integers?

(1) n^5∗n^(−2) has exactly 5 divisors between 1 and itself, both not inclusive.

(2) 3*n^2 has exactly 8 divisors between 1 and itself, both not inclusive.

st1) n^3 has 7 divisors (all inclusive). Thus n^3 can't be factorized to multiple prime numbers. n^3 can be represented as x^6 where x is a prime number.

Hence, n = x^2 where x is a prime number and factors of n = 3 SUFFICIENT

st2) this is a bit tricky. 3*n^2 has total 10 divisors. So, n can either be a perfect square of a prime number (3x^4) or 3*3^8. In such cases, factors of n can be either 3 or 5. So INSUFFICIENT.

The answer should be  A

hiranmay · Answer

If n is a positive integer, then n is a multiple of how many positive integers?

(1)  n^5*n^{-2}  has exactly 5 divisors between 1 and itself, both not inclusive. -->   suff:  n^5*n^{-2}  has 5 divisors excluding 1 & itself i.e.  n^3  has 5+2=7=1*7 divisors including 1 & itself, so  n^3 = p^6 , where p is a prime number => n= p^2. so n is a multiple of 2+1=3 +ve integers

(2)  3n^2  has exactly 8 divisors between 1 and itself, both not inclusive. -->   insuff:  3n^2  has 8 divisors excluding 1 & itself i.e.  3n^2  has 8+2=10=1*10=2*5 divisors including 1 & itself, so  3n^2 = a^9  or  3n^2 = b^1*c^4 , or  3n^2 = d^4*e^1 , where a,b,c,d,&e are prime numbers => if a=3, then  n^2 = a^8  or if b=3, then  n^2 = c^4 , or if d=3,  n^2 = d^3*e^1 => last one not possible, because n is a +ve integer, so  n = a^4  or  n = c^2 , so when  n = a^4 , n is a multiple of 4+1=5 +ve integers, but when  n = c^2 , n is a multiple of 2+1=3 +ve integers. so no definite answer from (2) 
  
 Answer: A

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MayankSingh · Answer

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   
  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

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HOT Competition: If n is a positive integer, then n is a multiple of

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HOT Competition: If n is a positive integer, then n is a multiple of

Prep Toolkit

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