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n, p and r

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n, p and r

by ayankm » Mon May 17, 2010 3:55 pm
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 answered as -E- but the actual answer says something else.
Any help would be much appreciated.
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by sk818020 » Mon May 17, 2010 4:50 pm
If n is odd then,

p + r =

even + odd = odd; or

odd + even = odd

because only an even plus an odd number will equal an odd number.

(1) Tells us that p and r are prime number. This tells us that either p or r must equal 2 because again;

n=odd=p+r

Thus, p+r must be = even + odd.

There is only 1 prime even number, which is 2. So p or r is 2. This does not tell us p is 2, so (1) is insufficient.

(2) tells us that r is not equal to 2. This does not tell us a whole lot. This tell us that if r is the even number r is not equal to 2. (2) is insufficient.

Put them together we know that either p or r must be equal to two because it must be an even and an odd prime number and the only prime even number is 2, and we know that r is not equal to 2. So p must be equal to 2.

IMO C

What is the OA?
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by ayankm » Mon May 17, 2010 5:01 pm
That is correct!!
I was not able to reason this properly in my mind.
Thanks for the explanation.
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by kstv » Mon May 17, 2010 9:19 pm
ayankm 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
sk818020 wrote
If n is odd then, p + r =
even + odd = odd; or
odd + even = odd
p and r both cannot be odd numbers, one of them has to be an even no.
2) r is not = 2, but it can be any other even no, then p is odd
if r is odd, then p is an even no but not necessarily 2. Insuff.
1) p and r are prime. One on them has to be 2 which is the only prime no which is even.
But which one ? . Insuff.
Combining if r is not 2, p has to be 2. Sufficient. IMO C.
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by sk818020 » Mon May 17, 2010 10:41 pm
kstv wrote:
ayankm 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
sk818020 wrote
If n is odd then, p + r =
even + odd = odd; or
odd + even = odd
p and r both cannot be odd numbers, one of them has to be an even no.
2) r is not = 2, but it can be any other even no, then p is odd
if r is odd, then p is an even no but not necessarily 2. Insuff.
1) p and r are prime. One on them has to be 2 which is the only prime no which is even.
But which one ? . Insuff.
Combining if r is not 2, p has to be 2. Sufficient. IMO C.
I did not say that p and r are both odd numbers. I was showing that one has to be odd and the other has to be even. I think that is evidenced throughout the rest of the discussion. If i need to clarify further, please let m know.

Thanks,

Jared
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