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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]  25 Jun 2005, 21:59
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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
Director
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Re: FG - DS [#permalink]  25 Jun 2005, 22:28
forrestgump wrote:
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

1) insuff, one of them has to equal 2 but it can be either of them
2) insuff

together suff r<>2 => p =2 C
Director
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What if n = 17?

Then if p and r are primes, combining both stmt 1 and 2,
r <> 2, then r + p = 17, 15 + 2 = 17
but then r(15) is not prime.

I say E.

Any thoughts?
VP
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C too
VP
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C too
Manager
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Folaa3 wrote:
What if n = 17?

Then if p and r are primes, combining both stmt 1 and 2,
r <> 2, then r + p = 17, 15 + 2 = 17
but then r(15) is not prime.

I say E.

Any thoughts?

Given:
N=P+R (from stem)
N = Odd (from stem)
Now, for N to be odd, either P or R shld be even and the other odd (E+O = O)
Also, P & R are prime (from S1)
Hence one of them shld be 2 (even prime)and the other an odd prime to satisfy the condn that N is odd.
To figure out which one of P & R is equal to 2, look at St2
St 2 : R <>2.
This implies that P has to be equal to 2
Hence 'C'
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Re: FG - DS [#permalink]  27 Jun 2005, 14:32
forrestgump wrote:
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

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

n is odd => p and r are (even,odd) or (odd,even)

S1 => p and r are prime. 2 is the only even Prime, so either p or r can be 2. But we don't know which is which. => Insufficient

S2 => r <> 2 => r can be 4 and p can be 5 or r can be 3 and p can be 2 etc. Infinite possibilities => Insufficient

S1 and S2 combined give us the fact that p=2.

C is my choice.
Re: FG - DS   [#permalink] 27 Jun 2005, 14:32
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