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>\ | + | 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>\boxed{\textbf{(D)}</math>. |
==See Also== | ==See Also== | ||
Revision as of 21:38, 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
![\[(n+1)(n+3)(n+5)(n+7)(n+9)\]](http://latex.artofproblemsolving.com/c/6/9/c694dd7f6739034188fe758646f1fd28508df784.png)
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 $\boxed{\textbf{(D)}$ (Error compiling LaTeX. Unknown error_msg).
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. 
