so are they asking us to count 1 or no

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so are they asking us to count 1 or no [#permalink]

19 Jul 2008, 15:58

so are they asking us to count 1 or no?

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:07

IMO A

Any number from 1 to p-1 does not have factor common with p, since the only common factor is 1.

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:07

Yes. 1 fits the definition.

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:10

If p is prime, then none of the number less than p can share a factor with p.

therefore A

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:16

yea but isnt 1 a factor and thus we cant count it as f(n)..

say n=3..Factors are 1 and 3..numbers less than 3 that are not factors of 3 are ..1 i.e 2..i didnt count 1 since 1 is already a factor of 3..

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:22

the question does state that "other than 1"

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 16:44

Well 1 is one of the acceptable factors. I'd include it
So f(5) = 4,3,2,1
A.

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Re: Gprep Prime number counting [#permalink]

19 Jul 2008, 17:34

fresinha12 wrote:

The question doesn't ask you to count how many numbers less than n are not divisors of n. It asks how many numbers less than n do not share any divisors (besides 1) with n, which is a different thing. 12 is not a divisor of 18, but 12 does share divisors with 18. In any case, you definitely must count one here: for any n>1, one will be an positive integer less than n which does not share any divisors with n (besides 1), and f(n) counts all such positive integers.
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Re: Gprep Prime number counting [#permalink] 19 Jul 2008, 17:34

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so are they asking us to count 1 or no

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so are they asking us to count 1 or no

so are they asking us to count 1 or no

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