2012 AMC 12B Problems/Problem 5


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?

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


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

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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