Difference between revisions of "2012 AMC 12B Problems/Problem 5"
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==Problem== | ==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? | + | Two integers have a sum of <math>26</math>. when two more integers are added to the first two, the sum is <math>41</math>. Finally, when two more integers are added to the sum of the previous <math>4</math> integers, the sum is <math>57</math>. What is the minimum number of even integers among the <math>6</math> integers? |
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | ||
Revision as of 06:20, 29 June 2023
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
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 |
| 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. 
