How many positive integers less than 300 have exactly 9 positive divis

Question

How many positive integers less than 300 have exactly 9 positive divisors?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

gmatophobia · Answer

If a integer, n, has 9 positive divisors, the prime factorization of the number will fall under one of the two cases shown below -

Case 1: n =  p^8  ; p is prime number

Case 2: n =  p^2 * q^2 ; p and q are prime numbers

We can try out the few prime numbers to obtain the numbers of integers that satisfy.

The detailed working is shown in the attached image.

Option C

VishvaT · Answer

Can someone explain me case 2? I dont know where does that come from. Or is there any reading material for the same.

gmatophobia · Answer

If  x = p_1^{n_1} * p_2^{n_2}*p_3^{n_3} ... ,

The number of factors =  (n_1+1)*(n_2+1)*(n_3+1)*....

Note:  p_1, p_2, p_3,..  represent the prime factors of x.

In this question, we are given that the number of factors is 9. There are two ways in which the product can be 9

Case 1 : 9 = 9 * 1 ⇒ In this case, only one prime number contributes to the total number of factor.

Hence, the number is of the form  x = p_1^{n_1}

n_1 + 1 = 9

n = 8

Case 2 : 9 = 3 * 3 ⇒ In this case, only two prime numbers contribute to the total number of factor.

Hence, the number is of the form  x = p_1^{n_1} * p_2^{n_2}

n_1 + 1 = n_2 + 1 = 3

n_1 = n_2 = 2

Bunuel · Answer

Finding the Number of Factors of an Integer

First, make the prime factorization of an integer  n = a^p * b^q * c^r , where  a ,  b , and  c  are prime factors of  n , and  p ,  q , and  r  are their respective powers.

The number of factors of  n  will be expressed by the formula  (p+1)(q+1)(r+1) .  NOTE:  this will include 1 and  n  itself.

Example:  Finding the number of all factors of 450:  450 = 2^1 * 3^2 * 5^2

The total number of factors of 450, including 1 and 450 itself, is  (1+1)(2+1)(2+1) = 2*3*3 = 18  factors.

According to the above, a positive integer to have 9 factors its prime factorization has to be either  n = p^8  or  n = p^2q^2 , where p and q are prime numbers - in this case, the number of factors will be (8 + 1) and (2 + 1)(2 + 1) = 9. Those are the only ways to get the product of 9.

p^8 < 300  is true only for one value, p = 2.

n=p^2q^2 < 300  is true for four pairs: (2, 3), (2, 5), (2, 7), and (3, 5).

Therefore, there are a total of 5 such numbers.

Answer: C.

How many positive integers less than 300 have exactly 9 positive divis

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GMAT Problem Solving (PS) Questions

How many positive integers less than 300 have exactly 9 positive divis

Prep Toolkit

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