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M23 #35

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M23 #35 [#permalink] New post 29 Aug 2008, 08:59
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

[Reveal] Spoiler: OA
B

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Re: m23 # 35 [#permalink] New post 29 Aug 2008, 10:19
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... x-1

there are x-1 integers that are less than x
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Re: m23 # 35 [#permalink] New post 29 Aug 2008, 14:48
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] New post 04 Aug 2009, 18:47
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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) = x-2 numbers
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Re: M23 #35 [#permalink] New post 05 Aug 2009, 16:48
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) = x-2 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 x-1.
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Re: M23 #35 [#permalink] New post 05 Aug 2009, 16:54
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) = x-2 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 x-1.



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] New post 05 Aug 2009, 17:04
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) = x-2 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 x-1.



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] New post 05 Aug 2009, 17:39
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 8-)
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Re: M23 #35 [#permalink] New post 10 Mar 2010, 19:19
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) = x-2 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) = x-1 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] New post 12 Mar 2010, 19:25
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] New post 12 Mar 2010, 20:54
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I also fell for the wording and chose x-2 (A), but now I also agree that the ans should be x-1 or B.
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Re: M23 #35 [#permalink] New post 13 Mar 2010, 03:05
yup, (x-2) was confusing until I read the question again and figured out that I could use 1.
(x-1) it is.
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Re: M23 #35 [#permalink] New post 11 Sep 2010, 22:03
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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

[Reveal] Spoiler: OA
B

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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 x-1 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 --> 7-1=6.

Answer: B.
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Re: M23 #35 [#permalink] New post 14 Mar 2011, 05:55
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] New post 14 Mar 2011, 20:18
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) = x-2 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 x-1.



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] New post 15 Mar 2011, 07:47
b; x-1, based on the definition of a prime number and the requirement of positive integers
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Re: M23 #35 [#permalink] New post 16 Apr 2012, 11:59
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 (prime-1) positive integers less than this prime that don't share factors. (B)
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Re: M23 #35 [#permalink] New post 19 Mar 2013, 05:34
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 x-1
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If x is a positive integer, f(x) is defined [#permalink] New post 12 Aug 2013, 22:06
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] New post 12 Aug 2013, 23:30
Hi,

If x is prime for example x = 5,
then f(x) will consist of (2, 3, 4) or x-2 integers

Thus, (A)

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Re: If x is a positive integer, f(x) is defined   [#permalink] 12 Aug 2013, 23:30
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