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
Number Properties
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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?
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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Hi Folks,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
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