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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==
 
==Solution==
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<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>
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<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 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 <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>
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2011|num-b=4|num-a=6|ab=B}}
 
{{AMC12 box|year=2011|num-b=4|num-a=6|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:40, 7 August 2023

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 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 $60$, so the second smallest multiple of $60$ is $2\times60=120$. Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$

See also

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

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