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Math Revolution GMAT Instructor
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For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]
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01 Jan 2018, 02:34
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70% (01:01) correct 30% (01:09) wrong based on 61 sessions
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[GMAT math practice question] For positive integer n, with distinct prime factors p1, p2,…,pn, the function \(f(n) = n(1\frac{1}{p1})(1\frac{1}{p2})(1\frac{1}{p3})\)\(….(1 \frac{1}{pk})\) gives the number of positive integers less than n which have no common factor with n except 1. What is the value of f(30) ? A. 5 B. 6 C. 7 D. 8 E. 9
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For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]
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01 Jan 2018, 04:14
MathRevolution wrote: [GMAT math practice question]
For positive integer n, with distinct prime factors p1, p2,…,pn, the function \(f(n) = n(1\frac{1}{p1})(1\frac{1}{p2})(1\frac{1}{p3})\)\(….(1 \frac{1}{pk})\) gives the number of positive integers less than n which have no common factor with n except 1. What is the value of f(30) ? A. 5 B. 6 C. 7 D. 8 E. 9 Prime factors of \(30=2*3*5\) So \(f(30)=30*(1\frac{1}{2})*(1\frac{1}{3})*(1\frac{1}{5})=8\) Option D



Math Revolution GMAT Instructor
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Posts: 5839
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Re: For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]
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03 Jan 2018, 01:41
=> Since \(30 = 2*3*5\),\(f(30) = 30*(1\frac{1}{2})(1\frac{1}{3})(1\frac{1}{5})\)\(= 30*(\frac{1}{2})(\frac{2}{3})(\frac{4}{5})=8.\) Therefore, the answer is D. Answer : D
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Re: For positive integer n, with distinct prime factors p1, p2,…,pn, the f
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03 Jan 2018, 01:41






