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M23-35

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New post 16 Sep 2014, 01:20
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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
[Reveal] Spoiler: OA

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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.


Answer: B
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M23-35 [#permalink]

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New post 15 Jun 2015, 06:33
Hi Bunuel,

Please correct my understanding here.

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.


Answer: B

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New post 15 Jun 2015, 06:43
svijayaug12 wrote:
Hi Bunuel,

Please correct my understanding here.

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.


Answer: B


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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Re: M23-35 [#permalink]

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New post 28 Jul 2015, 08:17
I chose A for same reason.
Question doesnt seem very clear whether 1 should be included or not.

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New post 28 Jul 2015, 08:26
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M23-35 [#permalink]

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New post 31 Jul 2015, 09:27
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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New post 10 Jan 2016, 17:57
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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New post 01 Apr 2016, 04:13
I think this is a poor-quality question and I agree with explanation.

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Re: M23-35 [#permalink]

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New post 21 Oct 2016, 21:22
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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Re: M23-35 [#permalink]

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New post 14 Aug 2017, 05:21
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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Re: M23-35   [#permalink] 14 Aug 2017, 05:21
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