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Bunuel
If p and q are integers and 20p + 3q is odd, which of the following must be odd?

(A) p – q

(B) p + 2q

(C) 3p + q

(D) 2p + q^2

(E) 3p + 3q

There are three rules that are useful for this problem:

even x any integer = even

even + odd = odd

odd x odd = odd

We see that 20p must be even, so we have:

even + 3q = odd

We see that 3q must be odd, so q must be odd.

Therefore we see that answer D must be odd, because 2p is even and q^2 = odd x odd = odd.

Answer: D
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If p and q are integers and 20p + 3q is odd, which of the following must be odd?

According to the question, we can infer that q = odd and nothing can be said about p i.e. p can be odd or even

Going by the options

(A) p – q = even/odd - odd = even/odd. Incorrect

(B) p + 2q = even/odd + even = even/odd. Incorrect

(C) 3p + q = even/odd + odd = even/odd. Incorrect

(D) 2p + q^2 = even + odd = odd. correct hence answer

(E) 3p + 3q = even/odd + odd = even/odd. Incorrect
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