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>{\boxed{\textbf{(D)15}}}</math> is the correct answer.
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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{(D)15}}</math> is the correct answer.
  
 
==See Also==
 
==See Also==

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 ${\boxed{(D)15}}$ 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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