Difference between revisions of "2003 AMC 12B Problems/Problem 12"
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== Solution == | == Solution == | ||
− | + | For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is <math>3 \cdot 5 = 15</math>, so <math>\framebox{D}</math>. | |
==See Also== | ==See Also== |
Revision as of 22:36, 4 January 2017
- The following problem is from both the 2003 AMC 12B #12 and 2003 AMC 10B #18, so both problems redirect to this page.
Problem
What is the largest integer that is a divisor of
for all positive even integers ?
Solution
For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is , so
.
See Also
2003 AMC 12B (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 |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.