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? |
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<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> | ||
==Solution== | ==Solution== | ||
− | + | Since, <math>x + y = 26</math>, <math>x</math> can equal <math>15</math>, and <math>y</math> can equal <math>11</math>, so no even integers are required to make 26. To get to <math>41</math>, we have to add <math>41 - 26 = 15</math>. If <math>a+b=15</math>, at least one of <math>a</math> and <math>b</math> must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from <math>26</math> to <math>41</math>. Finally, we have the last transition is <math>57-41=16</math>. If <math>m+n=16</math>, <math>m</math> and <math>n</math> can both be odd because two odd numbers sum to an even number, meaning only <math>1</math> even integer is required. The answer is <math>\boxed{\textbf{(A)}}</math>. ~Extremelysupercooldude (Latex, grammar, and solution edits) | |
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+ | ==Solution 2== | ||
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+ | 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. | ||
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+ | 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. | ||
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+ | Finally, <math>57-41=16</math>, and once again, <math>16</math> is an even number so both of these integers can be odd. | ||
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+ | Therefore, we have a total of one even integer and our answer is <math>\boxed{\textbf{(A)}}</math>. | ||
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== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2012|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:28, 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
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 |
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