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Functions

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Manager
Manager
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Joined: 14 Dec 2008
Posts: 171
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Kudos [?]: 14 [0], given: 39

Functions [#permalink] New post 29 Sep 2009, 11:12
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

75% (01:52) correct 25% (01:48) wrong based on 28 sessions
The function f is defined for all +ve integers n by the following rule: f(n) is the number of +ve intergers each of which is less than n and has no +ve 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
Manager
Manager
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Joined: 24 Aug 2009
Posts: 151
Followers: 3

Kudos [?]: 33 [0], given: 46

Re: Functions [#permalink] New post 29 Sep 2009, 12:01
manojgmat wrote:
The function f is defined for all +ve integers n by the following rule: f(n) is the number of +ve intergers each of which is less than n and has no +ve 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


we need to solve it by picking numbers, i picked 5, 7 , 11 and 23 , all satisfies P-1 , hence A. Because since N is prime , it is divisible only by itself and 1, so there will be no common positive factors other than 1 for all +ve integers less than N. hence P-1
Senior Manager
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Joined: 31 Aug 2009
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Location: Sydney, Australia
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Kudos [?]: 115 [0], given: 20

Re: Functions [#permalink] New post 30 Sep 2009, 20:48
Since p is prime, by definition it only has as its +ve factors {P,1}. Therefore nothing below N will ever share a common factor with it aside from 1. Answer will always be p-1
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Re: Functions [#permalink] New post 02 May 2011, 22:28
for 2 , it is 1
for 3, 1,2 hence 2
for 5, 1,2,3,4 hence 4
for 7, 1,2,3,4,5,6 hence 6

thus p-1
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Re: Functions   [#permalink] 02 May 2011, 22:28
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