Difference between revisions of "2011 AMC 12B Problems/Problem 5"

Revision as of 10:02, 4 July 2013

Problem

Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$ . What is the sum of the digits of $N$ ?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution

$N$ must be divisible by every positive integer less than $7$ , or $1, 2, 3, 4, 5,$ and $6$ . Each number that is divisible by each of these is is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$ , so each number divisible by these is a multiple of $60$ . The smallest multiple of $60$ is clearly $60$ , so the second smallest multiple of $60$ is $2\times60=120$ . Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 ==See also==
 {{AMC12 box|year=2011|num-b=4|num-a=6|ab=B}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2011 AMC 12B Problems/Problem 5"

Revision as of 10:02, 4 July 2013

Problem

Solution

See also