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

So, x+y=26, x could equal 15, and y could equal 11, so no even integers required here. 41-26=15. a+b=15, a could equal 9 and b could equal 6, so one even integer is required here. 57-41=16. m+n=16, m could equal 9 and n could equal 7, so no even integers required here, meaning only 1 even integer is required; A.

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 $\boxed{(\text{A})}$ .

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2012 AMC 12B Problems/Problem 5

Contents

Problem

Solution

Solution 2

See Also