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If n = p + r, where n, p, and r are positive integers and n

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Manager
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If n = p + r, where n, p, and r are positive integers and n [#permalink] New post 06 May 2008, 09:50
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If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
(1) p and r are prime numbers.
(2) r ≠ 2

i know that neither st.1 nor st.2 is insuff. but i can't come up with example. can anyone tell me example of st. 1?
Manager
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Re: DS 6-23 [#permalink] New post 06 May 2008, 10:16
rewrite question:
n-r=p

1)
3-2=1
5-3=2
insufficient

2) n is odd
odd
3-1=2
11-3=8
insufficient.

1&2)
all primes except 2 are odd, thus r is odd. The problem states n is odd.
odd-odd=even
therefore p must be equal to 2 to be even and a prime number.

C
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Re: DS 6-23 [#permalink] New post 06 May 2008, 10:36
C for me as well.

Since n is odd, therefore either p is odd and r even, or p even and r odd. I worked with stat 2 first.

Stat 2 says that r is not 2. Thats fine, r can be anything else, as can s; for example r=1 and s=4. Insuff.

Stat 1 says that p and r are prime. So we know that immediately p or r is 2, because all other primes are odds and odd+odd=even. But, either p can be 2 or r can be 2. Insuff.

Together, we have all we need.
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Re: DS 6-23 [#permalink] New post 07 May 2008, 04:52
C for me

If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?

(1) p and r are prime numbers. -> n is odd thus of P,R one has to be even n the other odd
now we don't know which is even,which odd thus insuff.
(2) r ≠ 2 = does not provide information -> insuff

comb 1& 2 p&r are prime and one has to be even and the other odd, using stmt-2 r is not 2 thus p=2 thus C
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Re: DS 6-23   [#permalink] 07 May 2008, 04:52
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If n = p + r, where n, p, and r are positive integers and n

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