2003 AMC 12B Problems/Problem 12: Difference between revisions
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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]]}} | ||
==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 23:57, 4 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
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 | ||
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