2012 AMC 12B Problems/Problem 5
Two integers have a sum of . when two more integers are added to the first two, the sum is . Finally, when two more integers are added to the sum of the previous integers, the sum is . What is the minimum number of even integers among the integers?
Since, , can equal , and can equal , so no even integers are required to make 26. To get to , we have to add . If , at least one of and must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from to . Finally, we have the last transition is . If , and can both be odd because two odd numbers sum to an even number, meaning only even integer is required. The answer is . ~Extremelysupercooldude (Latex, grammar, and solution edits)
Just worded and formatted a little differently than above.
The first two integers sum up to . Since 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 . Since is odd, we must have at least one even integer in these next two.
Finally, , and once again, 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 .
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