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

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abhi758
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If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   08 Apr 2010, 03:56
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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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C
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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   08 Apr 2010, 04:05
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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: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   08 Apr 2010, 11:29
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: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   09 Apr 2010, 14:06
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
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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   17 Dec 2015, 07:35
If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
for n to be odd, p must be odd and r even or p even and r odd.


(1) p and r are prime numbers.
it might be that p=2 and r= some other prime number, different than 2, since n must be odd.
or r=2, and p an odd prime number. thus, statement 1 is insufficient.

(2) r ≠ 2
statement 2 alone is insufficient.

1+2
r is not 2, thus r is an odd prime number.
this means that p is 2.

1+2 is sufficient. C
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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   20 Dec 2015, 05:03
Odd +(-) Odd= Even
Even +(-) Even= Even
Odd + Even = Odd
Even + Odd = Even

n=p+r
and n is odd

So, either p or r is odd.

Statement 1. Although it indicates one of them is 2 (the only even prime number), it does not clarify whether its p or r. Insuff.
Statement 2. Numerous even and odd combinations would give odd. Insuff.

Combining the two, the air is clear in terms of which is not 2. R is not 2. That implies P is 2.

Hence, C
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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink] New post   15 Apr 2024, 12:48
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Re: If n = p + r, where n, p, and r are positive integers and n [#permalink]
15 Apr 2024, 12:48
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