Art of Problem Solving
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2003 AMC 12B Problems/Problem 12: Difference between revisions

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What is the largest integer that is a divisor of <math>(n+1)(n+3)(n+5)(n+7)(n+9)</math> for all positive even integers <math>n</math>?
{{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>
Line 7: Line 15:
== 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

\[(n+1)(n+3)(n+5)(n+7)(n+9)\]

for all positive even integers $n$?

$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$

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 $3 * 5 = 15$, so $\framebox{D}$.

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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