2012 AMC 12B Problems/Problem 5

Problem

Two integers have a sum of $26$. when two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

Since, $x + y = 26$, $x$ can equal $15$, and $y$ can equal $11$, so no even integers are required to make 26. To get to $41$, we have to add $41 - 26 = 15$. If $a+b=15$, at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$. Finally, we have the last transition is $57-41=16$. If $m+n=16$, $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\boxed{\textbf{(A)}}$. ~Extremelysupercooldude (Latex, grammar, and solution edits)

Solution 2

Just worded and formatted a little differently than above.

The first two integers sum up to $26$. Since $26$ is even, in order to minimize the number of even integers, we make both of the first two odd.

The second two integers sum up to $41-26=15$. Since $15$ is odd, we must have at least one even integer in these next two.

Finally, $57-41=16$, and once again, $16$ is an even number so both of these integers can be odd.

Therefore, we have a total of one even integer and our answer is $\boxed{\textbf{(A)}}$.