Difference between revisions of "2012 AMC 10B Problems/Problem 9"
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| − | Out of the first two integers, it's possible for both to be | + | Out of the first two integers, it's possible for both to be odd: for example, <math>11 + 15 = 26.</math> But the next two integers, when added, increase the sum by <math>15,</math> which is odd, so one of them must be odd and the other must be even: for example, <math>3 + 12 = 15.</math> Finally, the next two integers increase the sum by <math>16,</math> which is even, so we can have both be odd: for example, <math>9 + 7 = 16.</math> Therefore, <math>\boxed{\textbf{(A) } 1}</math> is the minimum number of integers that must be even. |
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Revision as of 16:13, 19 January 2014
Problem 9
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of even integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\5$ (Error compiling LaTeX. Unknown error_msg)
Solution
Out of the first two integers, it's possible for both to be odd: for example, 
But the next two integers, when added, increase the sum by 
which is odd, so one of them must be odd and the other must be even: for example, 
Finally, the next two integers increase the sum by 
which is even, so we can have both be odd: for example, 
Therefore, 
is the minimum number of integers that must be even.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
