Difference between revisions of "2012 AMC 12B Problems/Problem 5"
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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 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? | ||
| − | + | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | |
==Solution== | ==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. | 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 <math>26</math>. Since <math>26</math> 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 <math>41-26=15</math>. Since <math>15</math> is odd, we must have at least one even integer in these next two. | ||
| + | |||
| + | Finally, <math>57-41=16</math>, and once again, <math>16</math> 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 <math>\boxed{(\text{A})}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC12 box|year=2012|ab=B|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Revision as of 00:00, 17 February 2016
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 |
| 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. 
