2012 AMC 12B Problems/Problem 5
Contents
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?
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 . 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 |
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All AMC 12 Problems and Solutions |
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