2012 AMC 12B Problems/Problem 5
Contents
Problem
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?
Solution
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)
Solution 2
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 .
See Also
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.