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Re M2335
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16 Sep 2014, 01:20
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 \(x1\) 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 \(71=6\) numbers. Answer: B
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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) > x2. eg: say prime number is 7. Numbers less than 7 other than 1 > 2, 3,4,5,6. Total  5 and hence X2. 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 \(x1\) 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 \(71=6\) numbers.
Answer: B



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Re: M2335
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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) > x2. eg: say prime number is 7. Numbers less than 7 other than 1 > 2, 3,4,5,6. Total  5 and hence X2. 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 \(x1\) 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 \(71=6\) 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: M2335
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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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28 Jul 2015, 08:26



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Re: M2335
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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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Re M2335
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10 Jan 2016, 17:57
I think this is a poorquality question and I agree with explanation. This doesn't seem like a 700 level question.



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Re M2335
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01 Apr 2016, 04:13
I think this is a poorquality question and I agree with explanation.



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Re: M2335
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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 1The 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 1Hence, 1 has to be in the set! Total elements: 6 \(x1\) Option B
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Re: M2335
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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 M2335
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09 Nov 2017, 18:17
I think this is a poorquality question and I agree with explanation.



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Re M2335
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06 Dec 2017, 18:34
I think this is a highquality question and I don't agree with the explanation. Isn't 1 also to be excluded from this, making it x2 . Please elaborate



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Re: M2335
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06 Dec 2017, 20:42



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Re: M2335
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13 Mar 2018, 18:33
I missed to include 1 and selected x2 as answer which was incorrect. Thanks bunuel for the explanation on why we should consider 1 as well. Ans: x1
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Re M2335
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23 Mar 2018, 23:57
I think this the explanation isn't clear enough, please elaborate. Why is 1 counted with other numbers?



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Re: M2335
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24 Mar 2018, 00:49



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Re: M2335
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03 Apr 2018, 03:41
A tricky one !!! The answer is option B. Key learning  read each and every word with utmost care !!!
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Re: M2335
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03 Apr 2018, 10:39
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










