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GRE Prep Club Team Member
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x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that
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24 Mar 2021, 05:00
24 Mar 2021, 05:00
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\(x = 9^{10}– 3^{17}\) and \(\frac{x}{n}\) is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five
(A) One
(B) Two
(C) Three
(D) Four
(E) Five
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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that
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24 Mar 2021, 08:56
24 Mar 2021, 08:56
1
GeminiHeat wrote:
\(x = 9^{10}– 3^{17}\) and \(\frac{x}{n}\) is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five
(A) One
(B) Two
(C) Three
(D) Four
(E) Five
\(x = 9^{10}– 3^{17}\)
\(x = 3^{20}– 3^{17}\)
\(x = 3^{17}(3^3 – 1)\)
\(x = 3^{17}(27 – 1)\)
\(x = 3^{17}(2)(13)\)
\(n\) is a positive integer that has exactly two factors - means \(n\) is a prime number
So, \(n\) can have 3 values - \(2\), \(3\) and \(13\)
Hence, option C
Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that
[#permalink]
24 Mar 2021, 09:02
24 Mar 2021, 09:02
1
x= 9^10 – 3^17
x= 3^2*10 - 3^17
x= 3^20 -3^17
x= 3^17 ( 3^3 - 1)
x= 3^17 (27-1)
x = 3^17 (26)
x= 3^17 (2) (13)
since the question says n is a positive integer that has exactly two factors, that means it is a prime number
Hence n will have 3 values - 3, 2, 13
Answer - C
x= 3^2*10 - 3^17
x= 3^20 -3^17
x= 3^17 ( 3^3 - 1)
x= 3^17 (27-1)
x = 3^17 (26)
x= 3^17 (2) (13)
since the question says n is a positive integer that has exactly two factors, that means it is a prime number
Hence n will have 3 values - 3, 2, 13
Answer - C
