Difference between revisions of "2003 AMC 12B Problems/Problem 12"
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| − | What is the largest integer that is a divisor of < | + | {{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #12]] and [[2003 AMC 10B Problems|2003 AMC 10B #18]]}} |
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| + | ==Problem== | ||
| + | |||
| + | What is the largest integer that is a divisor of | ||
| + | |||
| + | <cmath>(n+1)(n+3)(n+5)(n+7)(n+9)</cmath> | ||
| + | |||
| + | for all positive even integers <math>n</math>? | ||
<math> | <math> | ||
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== Solution == | == Solution == | ||
Since 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 * 5 = 15</math>, so <math>\framebox{D}</math>. | Since 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 * 5 = 15</math>, so <math>\framebox{D}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2003|ab=B|num-b=11|num-a=13}} | ||
| + | {{AMC10 box|year=2003|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:57, 5 January 2014
- 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
Since 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. 
