Difference between revisions of "2003 AMC 12B Problems/Problem 12"
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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>. | ||
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Revision as of 09:25, 4 July 2013
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 
.
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