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

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How many positive integers less than 300 have exactly 9 positive divis [#permalink]
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VishvaT wrote:
Can someone explain me case 2? I dont know where does that come from. Or is there any reading material for the same.

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 $$x = p_1^{n_1} * p_2^{n_2}$$

$$n_1 + 1 = n_2 + 1 = 3$$

$$n_1 = n_2 = 2$$
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Re: How many positive integers less than 300 have exactly 9 positive divis [#permalink]

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

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

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

­

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.

Re: How many positive integers less than 300 have exactly 9 positive divis [#permalink]
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