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

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If n = p + r, where n, p, and r are positive integers and n [#permalink] New post 08 Apr 2010, 02:56
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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
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Re: Number Properties - Debatable OA [#permalink] New post 08 Apr 2010, 03:05
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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.

Answer: C.
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Re: Number Properties - Debatable OA [#permalink] New post 08 Apr 2010, 10:29
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

n can be odd only if a prime number is 2

1)
p and r can take values(2,3,5,7 etc)
if 2 is not there then the sum is not odd
also either p or r can be 2
hence insuffcient

2)
now r not equals 2 but we do not know what p is. both r and n can take any positive valeu to give n as odd

combined 1 and 2
since r not equals 2 and for n to be odd we need 2, p equals 2

hence C
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Re: Number Properties - Debatable OA [#permalink] New post 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
Re: Number Properties - Debatable OA   [#permalink] 09 Apr 2010, 13:06
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