Prac Exam #2 Fnction Problem

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Prac Exam #2 Fnction Problem

by fangtray » Sun Apr 22, 2012 4:04 am
The function f is defined for all positive integers by n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is any prime number then f(p) =

a. p-1
b. p-2
c. (p+1)/2
d. (p-1)/2
e. 2

no idea what the quesiton is asking.
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by khandelwal.ab » Sun Apr 22, 2012 4:19 am
This is pretty straight forward, pick a prime number..say 5.. now, we have to find all positive integers less than 5, which have no positive factor in common with 5 (other than 1). Since 5 is prime all numbers from 1 to 4 satisfy the condition.

Hence IMO: A

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by Shalabh's Quants » Sun Apr 22, 2012 6:14 am
fangtray wrote:The function f is defined for all positive integers by n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is any prime number then f(p) =

a. p-1
b. p-2
c. (p+1)/2
d. (p-1)/2
e. 2

no idea what the quesiton is asking.
Pl. see the solution here.

https://www.beatthegmat.com/gmat-prep-t73368.html
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by Anurag@Gurome » Sun Apr 22, 2012 5:30 pm
fangtray wrote:The function f is defined for all positive integers by n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is any prime number then f(p) =

a. p-1
b. p-2
c. (p+1)/2
d. (p-1)/2
e. 2

no idea what the quesiton is asking.
Tricky solution:

Let us take p = 2 (smallest prime)
Now number of positive integers less than p and has no common factor with p other than 1 is 1. So f(2) = 1

Only option A satisfies this result.

Mathematical Approach:

Note that a prime number will have common factors other than 1 only with its multiples like p², p³ etc. As p is always greater than 1, all multiples of p are greater than p. Hence, none of the integers less than p will have any common factor with p.

Thus, f(p) = Number of positive integers less than p = (p - 1)

The correct answer is A.
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