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# M23-35

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:20
1
6
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Difficulty:

55% (hard)

Question Stats:

59% (01:01) correct 41% (01:16) wrong based on 189 sessions

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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. $$\frac{(x + 1)}{2}$$
D. $$\frac{(x - 1)}{2}$$
E. 2

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16 Sep 2014, 00:20
2
Official Solution:

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. $$\frac{(x + 1)}{2}$$
D. $$\frac{(x - 1)}{2}$$
E. 2

Basically the 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 $$x$$).

For example consider $$x=7=prime$$: how many numbers are less than 7 and have no common factors with 7: 1, 2, 3, 4, 5, and 6, so total of $$7-1=6$$ numbers.

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15 Jun 2015, 05:33
Hi Bunuel,

Aren't we supposed to return the set of all the positive numbers less than x that does not have a common factor with x OTHER THAN 1.

In that case, aren't we supposed to ignore the positive factor 1 and find out how many other numbers less than x are there who do not have common factor with X.

With this as my understanding, I selected option (A) -> x-2.

eg: say prime number is 7.

Numbers less than 7 other than 1 -> 2, 3,4,5,6.
Total - 5 and hence X-2.

Bunuel wrote:
Official Solution:

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. $$\frac{(x + 1)}{2}$$
D. $$\frac{(x - 1)}{2}$$
E. 2

Basically the 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 $$x$$).

For example consider $$x=7=prime$$: how many numbers are less than 7 and have no common factors with 7: 1, 2, 3, 4, 5, and 6, so total of $$7-1=6$$ numbers.

Math Expert
Joined: 02 Sep 2009
Posts: 52902

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15 Jun 2015, 05:43
svijayaug12 wrote:
Hi Bunuel,

Aren't we supposed to return the set of all the positive numbers less than x that does not have a common factor with x OTHER THAN 1.

In that case, aren't we supposed to ignore the positive factor 1 and find out how many other numbers less than x are there who do not have common factor with X.

With this as my understanding, I selected option (A) -> x-2.

eg: say prime number is 7.

Numbers less than 7 other than 1 -> 2, 3,4,5,6.
Total - 5 and hence X-2.

Bunuel wrote:
Official Solution:

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. $$\frac{(x + 1)}{2}$$
D. $$\frac{(x - 1)}{2}$$
E. 2

Basically the 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 $$x$$).

For example consider $$x=7=prime$$: how many numbers are less than 7 and have no common factors with 7: 1, 2, 3, 4, 5, and 6, so total of $$7-1=6$$ numbers.

1 also does not have have a common factor with 7 other than 1. So, you should include 1 as well.
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28 Jul 2015, 07:17
I chose A for same reason.
Question doesnt seem very clear whether 1 should be included or not.
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28 Jul 2015, 07:26
rukna wrote:
I chose A for same reason.
Question doesnt seem very clear whether 1 should be included or not.

There is no reason whatsoever to exclude 1: 1 also does not have have a common factor with 7 other than 1. So, you should include 1 as well.
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31 Jul 2015, 08:27
Bunuel wrote:
rukna wrote:
I chose A for same reason.
Question doesnt seem very clear whether 1 should be included or not.

There is no reason whatsoever to exclude 1: 1 also does not have have a common factor with 7 other than 1. So, you should include 1 as well.

yeah, you are right. Got it.
Thanks Brunel
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GMAT 1: 600 Q33 V40
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10 Jan 2016, 16:57
I think this is a poor-quality question and I agree with explanation. This doesn't seem like a 700 level question.
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Location: Hong Kong
GMAT 1: 600 Q44 V28
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01 Apr 2016, 03:13
I think this is a poor-quality question and I agree with explanation.
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21 Oct 2016, 20:22
1
I think this is a very good GMAT like 'convoluted' question. The controversy related to 1 is actually because of the confusing language the question stem uses. Let me try and put it in another way:

...and do not have a common factor with $$x$$ other than 1

The above statement only says that 1 is a factor of numbers less than $$x$$ and not that 1 should be excluded from the set of number less than $$x$$

Lets take $$x = 7$$. Below is the set of $$x$$ according to question stem

6 - Less than 7, and does not have a common factor with 7 other than 1
5 - Less than 7, and does not have a common factor with 7 other than 1
4 - Less than 7, and does not have a common factor with 7 other than 1
3 - Less than 7, and does not have a common factor with 7 other than 1
2 - Less than 7, and does not have a common factor with 7 other than 1
1 - Less than 7, and does not have a common factor with 7 other than 1

Hence, 1 has to be in the set!

Total elements: 6
$$x-1$$

Option B
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14 Aug 2017, 04:21
First express the number N in its prime factors in the form:

N = a^p b^q c^r ...

where a, b, c, ... are different prime numbers and p, q, r, ... are
positive integers.

Then the number of positive integers less than N and prime to it is:

phi(N) = N(1 - 1/a)(1 - 1/b)(1 - 1/c).......

This is called Euler's function
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Joined: 17 Oct 2017
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09 Nov 2017, 17:17
I think this is a poor-quality question and I agree with explanation.
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Joined: 23 Feb 2017
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06 Dec 2017, 17:34
I think this is a high-quality question and I don't agree with the explanation. Isn't 1 also to be excluded from this, making it x-2 . Please elaborate
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06 Dec 2017, 19:42
rohan.0089 wrote:
I think this is a high-quality question and I don't agree with the explanation. Isn't 1 also to be excluded from this, making it x-2 . Please elaborate

This is explained above: There is no reason whatsoever to exclude 1: 1 also does not have have a common factor with 7 other than 1. So, you should include 1 as well.
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13 Mar 2018, 17:33
I missed to include 1 and selected x-2 as answer which was incorrect. Thanks bunuel for the explanation on why we should consider 1 as well.

Ans: x-1
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Joined: 23 May 2017
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23 Mar 2018, 22:57
I think this the explanation isn't clear enough, please elaborate. Why is 1 counted with other numbers?
Math Expert
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23 Mar 2018, 23:49
milindrastogi wrote:
I think this the explanation isn't clear enough, please elaborate. Why is 1 counted with other numbers?

This is explained in the discussion above.

1 also does not have have a common factor with 7 other than 1. So, you should include 1 as well.

For more you could read the thread above or check another discussion of this question here: https://gmatclub.com/forum/if-x-is-a-po ... ml#p938227

Hope it helps.
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03 Apr 2018, 02:41
A tricky one !!! The answer is option B. Key learning - read each and every word with utmost care !!!
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03 Apr 2018, 09:39
I misread this question and also thought the answer was A, because we were supposed to exclude 1. I think this was ambiguously worded and confusing, not actually difficult
Re: M23-35   [#permalink] 03 Apr 2018, 09:39
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# M23-35

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