Difference between revisions of "2003 AMC 12B Problems/Problem 12"

(Created page with "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>? <math> \text {(A) } 3 \qquad \text {(B) } ...")
 
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\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
 
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== Solution ==
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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>.

Revision as of 14:21, 1 July 2013

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 $\framebox{D}$.