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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>?
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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #12]] and [[2003 AMC 10B Problems|2003 AMC 10B #18]]}}
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==Problem==
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What is the largest integer that is a divisor of
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<cmath>(n+1)(n+3)(n+5)(n+7)(n+9)</cmath>
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for all positive even integers <math>n</math>?
  
 
<math>
 
<math>
 
\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165
 
\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165
 
</math>
 
</math>
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== Solution ==
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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)15}}}</math> is the correct answer.
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==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=11|num-a=13}}
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{{AMC10 box|year=2003|ab=B|num-b=17|num-a=19}}
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{{MAA Notice}}

Revision as of 22:40, 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

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

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

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 $3 \cdot 5 = 15$, so ${\boxed{\textbf{(D)15}}}$ is the correct answer.

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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