<p>If p,q,r,s, and t are whole numbers and the expression 2(p(q+r)+s)+t is even, which of the numbers must be even?</p>
<p>(A) p
(B) q
(C) r
(D) s
(E) t</p>
<p>2 * (x) will be even, whether or not x itself is even or odd. So the properties of p,q,r and s don't matter, for the purposes for this question.
So, 2(p(q+r)+s) + t = even_number + t</p>
<p>For this itself to be an even number, t must be an even number.</p>
<p>In other words, 2 times any integer is always even. So knowing that, screw what p,q,r, and s are. It doesn't matter. </p>
<p>Thus it must be t.</p>
<hr>
<p>Say 2(p(q+r)+s)=40
40+t=even, and thus t must be even, because if t were odd the result would be odd (say 40+3=43; odd)
Say 2(p(q+r)+s)=41 (odd)
41+t=even, and thus t must be even because no odd value of t could make the result even.</p>
<p>whatever is in parentheses is irrelevant. After all that nonsense in multiplied by 2, the result will be even. So you have even + t = even.
If t is odd, the result will be odd, but we know this isn't true. So, t must be even for the result to be even.</p>