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This topic is locked. If you want to discuss this question please repost it in the respective forum. If \(x\) is a positive integer, \(f(x)\) is defined as the number of positive integers which are less than \(x\) and do not have common factor with \(x\) other than 1. If \(x\) is prime then \(f(x) = ?\) (A) \(x  2\) (B) \(x  1\) (C) \(\frac{x + 1}{2}\) (D) \(\frac{x  1}{2}\) (E) 2 Source: GMAT Club Tests  hardest GMAT questions



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Re: M23 #35 [#permalink]
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11 Sep 2010, 23:03
sset009 wrote: If \(x\) is a positive integer, \(f(x)\) is defined as the number of positive integers which are less than \(x\) and do not have common factor with \(x\) other than 1. If \(x\) is prime then \(f(x) = ?\) (A) \(x  2\) (B) \(x  1\) (C) \(\frac{x + 1}{2}\) (D) \(\frac{x  1}{2}\) (E) 2 Source: GMAT Club Tests  hardest GMAT questions The confusing moment in this question is its wording. Basically question is: how many positive integers are less than given prime number x which has no common factor with x except 1. Well as x is a prime, all positive numbers less than x have no common factors with x (except common factor 1). So there would be x1 such numbers (as we are looking number of integers less than x). If we consider x=7 how many numbers are less than 7 having no common factors with 7: 1, 2, 3, 4, 5, 6 > 71=6. Answer: B.
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Re: M23 #35 [#permalink]
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04 Aug 2009, 19:47
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The answer is wrong. OA should be (A).
f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers



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Re: M23 #35 [#permalink]
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12 Mar 2010, 21:54
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I also fell for the wording and chose x2 (A), but now I also agree that the ans should be x1 or B.



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Re: m23 # 35 [#permalink]
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29 Aug 2008, 11:19
sset009 wrote: i disagree with teh OA. just want to get an opinion I will go with B. X is prime number.. value <x are 1.2... x1 there are x1 integers that are less than x
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Re: m23 # 35 [#permalink]
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29 Aug 2008, 15:48
i was thinking of x = 3
the number of numbers below three that do not have a any factor apart from 1 is 2 since 1 has a factor of 1 itself, it doenst count. therefore, \(f(x3)\) = 1
similarily, \(f(2)\) = 0



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Re: M23 #35 [#permalink]
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05 Aug 2009, 17:48
bipolarbear wrote: The answer is wrong. OA should be (A).
f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers bipolar u're forgetting 1. u need to count 1 as well. hence f(2) = 1 f(3) = 1,2 > 2 f(4) = 1,2,3 > 3 i got this exact question on the gmatprep and answer is x1.



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Re: M23 #35 [#permalink]
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05 Aug 2009, 17:54
viperm5 wrote: bipolarbear wrote: The answer is wrong. OA should be (A).
f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers bipolar u're forgetting 1. u need to count 1 as well. hence f(2) = 1 f(3) = 1,2 > 2 f(4) = 1,2,3 > 3 i got this exact question on the gmatprep and answer is x1. but the question states f(x) = number of factors OTHER than 1, so you can't count one... but i guess if gmatprep says so am i just interpreting the question wrong?



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Re: M23 #35 [#permalink]
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05 Aug 2009, 18:04
bipolarbear wrote: viperm5 wrote: bipolarbear wrote: The answer is wrong. OA should be (A).
f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers bipolar u're forgetting 1. u need to count 1 as well. hence f(2) = 1 f(3) = 1,2 > 2 f(4) = 1,2,3 > 3 i got this exact question on the gmatprep and answer is x1. but the question states f(x) = number of factors OTHER than 1, so you can't count one... but i guess if gmatprep says so am i just interpreting the question wrong? i fell for the wording too f(x) defined as all numbers positive less than x AND do not have same factor with x other than 1....meaning that you DO include x..... do NOT have ...OTHER THAN 1  means you include 1.



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Re: M23 #35 [#permalink]
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05 Aug 2009, 18:39
ah i see it now... this is a tricky gmat problem and even if its part of gmatprep, i think its more of a verbal question than math



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Re: M23 #35 [#permalink]
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10 Mar 2010, 20:19
Ans:A f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers if Ans is b then f(2) = 1 = 1 numbers f(3) = 2 = 2 number f(5) = 4 = 4 numbers f(7) = 6 = 6 numbers f(x) = x1 numbers all numbers other then 1 is having comman factor 2 with x and it is given that x is do not have common factor with f(x)
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Re: M23 #35 [#permalink]
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12 Mar 2010, 20:25
Plug in 23 for x, f(x) = 22 Plug in another prime 37, f(x) = 36 Ans B
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Re: M23 #35 [#permalink]
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13 Mar 2010, 04:05
yup, (x2) was confusing until I read the question again and figured out that I could use 1. (x1) it is.
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Re: M23 #35 [#permalink]
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14 Mar 2011, 06:55
Quote: If x is a positive integer, f(x) is defined as the number of positive integers which are less than x and do not have common factor with x other than 1. If x is prime then f(x) = ? factors below 7 = 6,5,4,3,2, and 1 "6" has no common factor with "7" except digit "1" "5" has no common factor with "7" except digit "1" "4" has no common factor with "7" except digit "1" etc, etc By the same token, do we also say "1" has no common factor with "7" except itself?
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Re: M23 #35 [#permalink]
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14 Mar 2011, 21:18
viperm5 wrote: bipolarbear wrote: The answer is wrong. OA should be (A).
f(2) = 0 = 0 numbers f(3) = 2 = 1 number f(5) = 2,3,4 = 3 numbers f(7) = 2,3,4,5,6 = 5 numbers f(x) = x2 numbers bipolar u're forgetting 1. u need to count 1 as well. hence f(2) = 1 f(3) = 1,2 > 2 f(4) = 1,2,3 > 3 i got this exact question on the gmatprep and answer is x1. Agree that answer is B. A can't be answer because if we put 2 which is a prime number in place of X in option 1, we get 0, which is not a positive integer. All other options give fraction if we put prime numbers to check. Last option can't be for the obvious reasons. So, only option that satisfies the given condition is option B, which satisfies the condition for all the prime number. I don't agree with few people that we need to count 1 because anyhow 1 is not a prime number and in question it is clearly mentioned that x is a prime number. Thank!!!



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Re: M23 #35 [#permalink]
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15 Mar 2011, 08:47
b; x1, based on the definition of a prime number and the requirement of positive integers



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Re: M23 #35 [#permalink]
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16 Apr 2012, 12:59
1) Since this is a function of x when x is prime, the solution must be applicable to all primes.
2) If x is prime, then no other common factor other than 1 will be shared with any other integer less than x anyways.
Together, any prime has (prime1) positive integers less than this prime that don't share factors. (B)



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Re: M23 #35 [#permalink]
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19 Mar 2013, 06:34
all are positive integers and less than x ( including 1 ) and not multiples of x (naturally) .. x is prime . I.E this means f(x) = all integers from 0 to x not inclusive . This will be x1
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If x is a positive integer, f(x) is defined [#permalink]
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12 Aug 2013, 23:06
If x is a positive integer, f(x) is defined as the number of positive integers which are less than x and do not have a common factor with x other than 1. If x is prime, then f(x)=?
A)x−2 B)x−1 C)(x+1)2 D)(x−1)2 E)2



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Re: If x is a positive integer, f(x) is defined [#permalink]
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13 Aug 2013, 00:30
Hi,
If x is prime for example x = 5, then f(x) will consist of (2, 3, 4) or x2 integers
Thus, (A)
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Re: If x is a positive integer, f(x) is defined
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