How many positive divisors does integer m have?

Question

How many positive divisors does positive integer  m  have?

(1)  m 3 has exactly 13 positive divisors 
(2)  m 2 has exactly 9 positive divisors

chetan2u · Answer

Number of positive divisors of  a^xb^yc^z….   is (x+1)(y+1)(z+1), where a, b, c.. are prime numbers.

(1)  m ³ has exactly 13 positive divisors 
13, when written as product of integers, can be written as only 1*13
Say m =  a^xb^y… , then  m^3=a^{3x}b^{3y}  and divisors of  m^3  are (3x+1)(3y+1)..=1*13=(0+1)(12+1)
So  m^3=b^{12}=(b^{4})^3 . Thus, m= b^4  and number of divisors are 4+1 or 5.
Sufficient

(2)  m ² has exactly 9 positive divisors
9, when written as product of integers, can be written as 1*9 or 3*3
Say m =  a^xb^y… , then  m^2=a^{2x}b^{2y}  and divisors of  m^2  are (2x+1)(2y+1)..=1*9=(0+1)(8+1)
So  m^2=b^{8}=(b^{4})^2 . Thus, m= b^4  and number of divisors are 4+1 or 5.
But, when (2x+1)(2y+1)=3*3=(2+1)(2+1),  m^2=a^2b^2  and m=ab with number of divisors as (1+1)(1+1)=4
Insufficient

A

Archit3110 · Answer

a good question

target How many positive divisors does positive integer  m  have

point to note that it is only square of prime numbers which have odd number of factors

#1

m ³ has exactly 13 positive divisors

given m^3 has 13 +ve factors or divisors
here m has to be ( m^4)^3
so (m^4)^3 is the only possible case ; 
factors of m would be 5
sufficient
#2
 m ² has exactly 9 positive divisors
m has 9 factors
possible cases 
(m^4)^2 ; (m^2)^4 ; factors of m can be 5,3
also
m^2 =( a^2*b^2)
so ( a^2*b^2) ; m=a*b factors of m 4
insufficient
OPTION A is correct

SaquibHGMATWhiz · Answer

Solution:

We are asked the number of factors of positive integer m

if  m=(p_1)^a	imes (p_2)^b	imes (p_3)^c..... , where  p_1, p_2, p_3  and so on are distinct prime numbers, then total number of factors  =(a+1)	imes (b+1)	imes (c+1).....

Statement 1:    m^3  has exactly 13 positive divisors

Since 13 is a prime number, it can only be written as  13	imes 1  which when we compare with  (a+1)	imes (b+1) , we get  a=12  and  b=0

Thus, we can be sure that  m^3=p^{12}  where p is a prime number 
 ⇒m=p^{\frac{12}{3}} 
 ⇒m=p^4

So, total factors of  m=4+1=5

Thus,  statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2:    m^2  has exactly 9 positive divisors

9 can be written as either  9	imes 1  or  3	imes 3

in first case,  a+1=9  or  a=8  and  m^2=p^8  
In the second case,  a+1=3  or  a=2  and  b+1=3  or  b=2  and finally  m^2=(p_1)^2	imes (p_2)^2

In both these cases, we will get different total factors of m

Thus,  statement 2 alone is not sufficient

Hence the right answer is    Option A

How many positive divisors does integer m have?

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How many positive divisors does integer m have?

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