Difference between revisions of "2003 AMC 12A Problems/Problem 12"
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Therefore, the sum of the numbers on the middle three cards is <math>3+3+6=\boxed{\mathrm{(E)}\ 12}</math>. | Therefore, the sum of the numbers on the middle three cards is <math>3+3+6=\boxed{\mathrm{(E)}\ 12}</math>. | ||
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== See Also == | == See Also == |
Revision as of 03:25, 4 October 2020
- The following problem is from both the 2003 AMC 12A #12 and 2003 AMC 10A #24, so both problems redirect to this page.
Contents
Problem
Sally has five red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
Video Solution
https://www.youtube.com/watch?v=Q30Dt8q7vm4&feature=youtu.be
Solution
Let and designate the red card numbered and the blue card numbered , respectively.
is the only blue card that evenly divides, so must be at one end of the stack and must be the card next to it.
is the only other red card that evenly divides , so must be the other card next to .
is the only blue card that evenly divides, so must be at one end of the stack and must be the card next to it.
is the only other red card that evenly divides , so must be the other card next to .
doesn't evenly divide , so must be next to , must be next to , and must be in the middle.
This yields the following arrangement from top to bottom:
Therefore, the sum of the numbers on the middle three cards is .
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.