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For positive integer n, with distinct prime factors p1, p2,…,pn, the f

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For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]

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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
[Reveal] Spoiler: OA

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For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]

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New post 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
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Re: For positive integer n, with distinct prime factors p1, p2,…,pn, the f [#permalink]

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New post 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   [#permalink] 03 Jan 2018, 01:41
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