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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink]
08 Apr 2010, 03:05

Expert's post

abhi758 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

Given: \(odd=p+r\), where \(p\) and \(r\) are positive integers. Question is \(p=2\)?

(1) The sum of two prime can be odd only if one of them is 2 (if 2, the only even prime, is not one of the primes, then the sum of the two primes will be odd+odd=even). But we don't know which one equals to 2. Not sufficient.

(2) \(r\neq{2}\). Clearly insufficient.

(1)+(2) \(r\neq{2}\), so \(p\) must equal to \(2\). Sufficient.

Re: If n = p + r, where n, p, and r are positive integers and n [#permalink]
09 Apr 2010, 13:06

I agreed with both answers above.

It is easy to see that both S1 and S2 are insufficient.

When taken together, we know that n is odd, and r is a prime number greater than 2 such as 3,5,7,11,13, etc. For the sum of r and p to be odd, p must be even. The only even prime number is 2.

Hence C

gmatclubot

Re: If n = p + r, where n, p, and r are positive integers and n
[#permalink]
09 Apr 2010, 13:06

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