Difference between revisions of "2012 AMC 12B Problems/Problem 5"
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==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{ | + | 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) |
==Solution 2== | ==Solution 2== | ||
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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. | 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{ | + | Therefore, we have a total of one even integer and our answer is <math>\boxed{\textbf{(A)}}</math>. |
== See Also == | == See Also == | ||
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
