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The function f is defined for all the positive integers n by

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The function f is defined for all the positive integers n by [#permalink] New post 28 Oct 2010, 07:27
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The function f is defined for all the positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1 . if p is a prime number then f(p)?

A. p-1
B. p-2
C. (p+1)/2
D. (p-1)/2
E. 2
[Reveal] Spoiler: OA
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Re: the function f(n) [#permalink] New post 28 Oct 2010, 07:31
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anilnandyala wrote:
the function f is defined for all the positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1 . if p is a prime number then f(p)?
a p-1
b p-2
c (p+1)/2
d (p-1)/2
e 2

please provide ur answer with explanation


The confusing moment in this question is its wording. Basically question is: how many positive integers are less than given prime number p which has no common factor with p except 1.

Well as p is a prime, all positive numbers less than p have no common factors with p (except common factor 1). So there would be p-1 such numbers (as we are looking number of integers less than p).

If we consider p=7 how many numbers are less than 7 having no common factors with 7: 1, 2, 3, 4, 5, 6 --> 7-1=6.

Answer: A.
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Re: the function f(n) [#permalink] New post 09 Nov 2010, 06:46
The moment you put a prime number in the function f(n), notice that all the numbers lesser than n have no divisor clashing with divisor of n since n is prime!!.

For instance f(7)= {6, 5, 4, 3, 2, 1}

Thus for f(p) number of integers falling under this set will be p-1


Answer :- A
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Re: the function f(n) [#permalink] New post 09 Oct 2013, 08:36
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Okay let's see what happens here.

It took me 20 seconds to understand what the question was precisely asking for. But this is the most important step; do not attempt anything if you don't understand throughly the question.

What I figured is that the definition of "relatively prime" was pretty close to the description assigned to the question.

Two different numbers are said to be relatively prime whenever their GCF=1.

Let's pick a random example: how many numbers less than 105 are relatively prime to 105?

105=3(5)7 then the total number of relatively primes will be: 105(1-1/3)(1-1/5)(1-1/7)=48

Let's apply the same logic to our question and consider a random prime number: p(1-1/p)= p-1 which turns out to be the correct answer.
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Re: the function f(n)   [#permalink] 09 Oct 2013, 08:36
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