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Question of the Week- 11 ( If p is positive integer and p^2 has ..)

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Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post Updated on: 13 Aug 2018, 00:17
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e-GMAT Question of the Week #11


If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p\)?

Options

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24

To access all the questions: Question of the Week: Consolidated List

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Originally posted by EgmatQuantExpert on 11 Aug 2018, 10:57.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:17, edited 2 times in total.
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Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 11 Aug 2018, 19:57
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EgmatQuantExpert wrote:
e-GMAT Question of the Week #11


If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p\)?

Options

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24

To access all the QOW questions: Question of the Week: Consolidated List

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Number of factors =\(17=1*17=(16+1)\),
therefore \(p^2=a^16\) and \(p=a^8\), so p will have \((8+1) =9\) factors
Now \(p^3=(a^8)^3=a^24\), and factors are \((24+1)=25\)

Thus factors that do not divide p but divide \(p^3\) are 2\(5-9=16\)

A
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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 11 Aug 2018, 22:33
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EgmatQuantExpert wrote:
e-GMAT Question of the Week #11


If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p\)?

Options

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24




Given \(p > 0\) & \(p^2\) has 17 factors, hence \(p\) is of the form \(n^8\) & has \((8+1) = 9\) factors

we get \(p^2 = (n^{8})^{2} = n^{16}\)

& \(p^3 = (n^{8})^{3} = n^{24}\), which has \((24+1)\) = \(25\) factors

Therefore # of factors of p^3 that are not factors of \(p = 25 - 9 = 16\)


Answer A.


Thanks,
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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 12 Aug 2018, 07:19
EgmatQuantExpert wrote:
e-GMAT Question of the Week #11


If \(p\) is a positive integer and \(p^2\) has total \(17\) positive factors, then find the number of positive integers that completely divides \(p^3\) but does not completely divide \(p\)?

Options

(A) 16
(B) 17
(C) 21
(D) 23
(E) 24

To access all the QOW questions: Question of the Week: Consolidated List

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As \(p^2\) has 17 factors, so \(p^2 = 2^{16}\) or \(p = 2^8\)........ (p can be any Prime Number)
Now, \(p^3 = 2^{24}\).. As per the question statement, the dividing Integer should divide \(2^{24}\) but not \(2^8\)...... So, the dividing Integer will be \(2^9, 2^{10},..... 2^{24}.\)

Hence, A.
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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 14 Aug 2018, 22:55
Can someone help me to elaborate more why you can conclude p^2 has 17 factors so p^2 = 2^16 or a^16. Sorry i dont get this point. Thank you so much
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Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 14 Aug 2018, 23:03
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Hungluu92vn wrote:
Can someone help me to elaborate more why you can conclude p^2 has 17 factors so p^2 = 2^16 or a^16. Sorry i dont get this point. Thank you so much


Total Number of Factors for \(a^{n} = n+1\).. Hope this is clear.
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Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 16 Aug 2018, 00:55

Solution



Given:
    • p is a positive integer
    • \(p^2\) has 17 positive factors

To find:
    • Number of positive integers that divides \(p^3\), but does not divide p

Approach and Working:
    • The number of factors of p^2 is 17
      o 17 is a prime number, so it can only be written as 1 * 17 = (0+1) * (16+1)
      o Thus, \(p^2 = P_1^{16}\), where \(P_1\) is a prime number
    • Implies, \(p = P_1^8\)
      o The number of factors of p will be 9
    • And, \(p^3 = P_1^{24}\)
      o The number of factors of \(p^3\) will be 25
    • All the factors of ‘p’ will also be the factors of \(p^3\)

Therefore, the number of positive integers that divides p^3, but does not divide p = 25 – 9 = 16

Hence, the correct answer is option A.

Answer: A

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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 16 Aug 2018, 01:14
Hungluu92vn wrote:
Can someone help me to elaborate more why you can conclude p^2 has 17 factors so p^2 = 2^16 or a^16. Sorry i dont get this point. Thank you so much


Since, \(p^2 = 17\), is a prime number,
• It can only be written as 1*17 = (0+1) * (16+1)
• And, we know that the number of factors of \(N = p^a * q^b * r^c\)... is (a+1) * (b+1) * (c+1) ...
Thus, \(p^2 = P_1^{16}\), where \(P_1\) is a prime number

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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)  [#permalink]

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New post 16 Aug 2018, 01:22
rahul16singh28 wrote:

As \(p^2\) has 17 factors, so \(p^2 = 2^{16}\) or \(p = 2^8\)........ (p can be any Prime Number)
Now, \(p^3 = 2^{24}\).. As per the question statement, the dividing Integer should divide \(2^{24}\) but not \(2^8\)...... So, the dividing Integer will be \(2^9, 2^{10},..... 2^{24}.\)

Hence, A.


There is a small correction in your solution.

The statement "p can be any Prime Number" is incorrect, because p is not a prime number,

Instead, it should have been stated as "p can have any prime number/factor in place of 2."
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Re: Question of the Week- 11 ( If p is positive integer and p^2 has ..)   [#permalink] 16 Aug 2018, 01:22
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