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Question of the Week 11 ( If p is positive integer and p^2 has ..)
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Updated on: 13 Aug 2018, 00:17
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eGMAT Question of the Week #11If \(p\) 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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Question of the Week 11 ( If p is positive integer and p^2 has ..)
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11 Aug 2018, 19:57
EgmatQuantExpert wrote: eGMAT Question of the Week #11If \(p\) 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\(59=16\) A
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Re: Question of the Week 11 ( If p is positive integer and p^2 has ..)
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11 Aug 2018, 22:33
EgmatQuantExpert wrote: eGMAT Question of the Week #11If \(p\) 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, GyM
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Re: Question of the Week 11 ( If p is positive integer and p^2 has ..)
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12 Aug 2018, 07:19
EgmatQuantExpert wrote: eGMAT Question of the Week #11If \(p\) 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 ..)
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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 ..)
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14 Aug 2018, 23:03
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 ..)
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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 ..)
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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 ..)
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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 ..)
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