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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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Updated on: 12 Aug 2018, 23:17
00:00

Difficulty:

75% (hard)

Question Stats:

45% (01:47) correct 55% (02:05) wrong based on 95 sessions

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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

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Originally posted by EgmatQuantExpert on 11 Aug 2018, 09:57.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:17, 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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11 Aug 2018, 18:57
2
1
EgmatQuantExpert wrote:
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

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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11 Aug 2018, 21:33
1
EgmatQuantExpert wrote:
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 $$p = 25 - 9 = 16$$

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

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12 Aug 2018, 06:19
EgmatQuantExpert wrote:
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

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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14 Aug 2018, 21:55
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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14 Aug 2018, 22:03
1
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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15 Aug 2018, 23:55

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.

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

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16 Aug 2018, 00:14
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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16 Aug 2018, 00:22
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 ..) &nbs [#permalink] 16 Aug 2018, 00:22
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