2011 AMC 12B Problem 5
Attempt Problem 5 of the 2011 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
5.
Let be the second smallest positive integer that is divisible by every positive integer less than What is the sum of the digits of
Answer: A
Solution:
A number divisible by every integer from to must be a multiple of
The second smallest positive multiple of is whose digit sum is
Thus, the correct answer is A.
Problem 5 in Other Years
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