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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]
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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.

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