Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 8 Problems/Problem 20"
Line 1: Line 1:
 +
==Problem==
 +
 
The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?
 
The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?
  

Revision as of 02:59, 16 January 2021

Problem

The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$

Solution

We wish to find possible values of $a$,$b$, and $c$. By finding the greatest common factor of $12$ and $15$, algebraically, it's some multiple of $b$ and from looking at the numbers, we are sure that it is 3, thus $b$ is 3. Moving on to $a$ and $c$, in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$, $\rightarrow 4$. Similarly with $3$ and $c$, we obtain $5$. The least common multiple of $4$ and $5$ is $20 \rightarrow \boxed{\textbf{(A)} 20}$

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_20&oldid=142205"