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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

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.

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.

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.
_________________

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.
_________________

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 _________________

One Kudos for an everlasting piece of knowledge is not a bad deal at all...

------------------------------------------------------------------------------------------------------------------------ Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover. -Mark Twain