Difference between revisions of "2011 AMC 12B Problems/Problem 5"
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math> | <math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math> | ||
| + | ==Solution== | ||
| + | <math>N</math> must be divisible by every positive integer less than <math>7</math>, or <math>1, 2, 3, 4, 5,</math> and <math>6</math>. Each number that is divisible by each of these is is a multiple of their least common multiple. <math>LCM(1,2,3,4,5,6)=60</math>, so each number divisible by these is a multiple of <math>60</math>. The smallest multiple of <math>60</math> is clearly <math>60</math>, so the second smallest multiple of <math>60</math> is <math>2\times60=120</math>. Therefore, the sum of the digits of <math>N</math> is <math>1+2+0=\boxed{3\ \textbf{(A)}}</math> | ||
Revision as of 13:35, 6 March 2011
Problem
Let 
be the second smallest positive integer that is divisible by every positive integer less than 
. What is the sum of the digits of ?
Solution
must be divisible by every positive integer less than , or 
and 
. Each number that is divisible by each of these is is a multiple of their least common multiple. 
, so each number divisible by these is a multiple of 
. The smallest multiple of is clearly , so the second smallest multiple of is 
. Therefore, the sum of the digits of is 
