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Number Properties

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Source: — Data Sufficiency |
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by MAAJ » Wed May 04, 2011 9:52 am
Posible combinations:

p * q-p * p = Even
O * E * O = Even
O * O * E = Even
E * O * O = Even
E * E * E = Even

(A) p -> Not necessarily
(B) q -> Not necessarily
(C) p-q -> Not necessarily
(D) pq -> Not necessarily
(E) p-pq -> Chose (E), but now that I'm testing (E), it looks like p-pq could be ODD or EVEN...

What's the OA?
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by clock60 » Wed May 04, 2011 11:33 am
hi koby gen
can you check the problem one more time
i as well as MAAJ also got that every option can be even or odd.
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by pankajks2010 » Thu May 05, 2011 7:04 pm
koby_gen wrote:If p and q are integers and p*q-p*p is even, which of the following must also be even ?

(A) p
(B) q
(C) p-q
(D) pq
(E) p-pq
Hi Folks,

For p*q-p*p to be even either both the terms (p*p and p*p) should be odd or both should be even.

Based on this the nature of p & q can be as follows:

a) When p is odd, q has to be odd
b) When p is even, q can be either odd or even

Now, using the above two, we can easily eliminate the first four choices.

Now, evaluating the two cases for the last option p-p*q:
a) When p is odd, then q also is odd thus, option E becomes (odd-odd=even)
b) When p is even, then q can be either odd/even. However, product of (even & odd) or (even & even) gives only an even number. Thus, the option E becomes (even-even=even)

Thus, the answer is E
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