M23 #35

sset009 · **Posted:** Fri Aug 29, 2008 8:59 am

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

Source: GMAT Club Tests - hardest GMAT questions

xyztroy · **Posted:** Fri Mar 12, 2010 8:54 pm

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.

sset009 · **Posted:** Fri Aug 29, 2008 8:59 am

i disagree with teh OA. just want to get an opinion

stallone · **Posted:** Fri Aug 29, 2008 10:13 am

x-2 for me !

x2suresh · **Posted:** Fri Aug 29, 2008 10:19 am

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

x2suresh · **Posted:** Fri Aug 29, 2008 10:20 am

stallone wrote:

x-2 for me !

Show me your work.. how did you get x-2

sset009 · **Posted:** Fri Aug 29, 2008 2:48 pm

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

bipolarbear · **Posted:** Tue Aug 04, 2009 6:47 pm

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

viperm5 · **Posted:** Wed Aug 05, 2009 4:48 pm

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.

bipolarbear · **Posted:** Wed Aug 05, 2009 4:54 pm

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?

viperm5 · **Posted:** Wed Aug 05, 2009 5:04 pm

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.

bipolarbear · **Posted:** Wed Aug 05, 2009 5:39 pm

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-)

dmetla · **Posted:** Wed Mar 10, 2010 11:48 am

Ans should be x-1

prabir2001 · **Posted:** Wed Mar 10, 2010 4:58 pm

Answer should be B.... Need to include '1'.

varun2410 · **Posted:** Wed Mar 10, 2010 7:19 pm

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)

ExecMBA2010 · **Posted:** Fri Mar 12, 2010 7:25 pm

Plug in 23 for x, f(x) = 22
Plug in another prime 37, f(x) = 36

Ans B

sidhu4u · **Posted:** Sat Mar 13, 2010 3:05 am

yup, (x-2) was confusing until I read the question again and figured out that I could use 1.
(x-1) it is.

mailpravs · **Posted:** Mon Sep 06, 2010 11:37 pm

B.

Bunuel · **Posted:** Sat Sep 11, 2010 10:03 pm

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

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 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.

gmatbull · **Posted:** Mon Mar 14, 2011 5:55 am

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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M23 #35

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