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 "2001 AMC 10 Problems/Problem 8"

(Created page with '== Problem == Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth sch…')
 
m (→‎Solution)
 
(8 intermediate revisions by 7 users not shown)
Line 3: Line 3:
 
Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
 
Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
  
−
<math> \textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252 </math>
+
<math> \textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252</math>
  
 
== Solution ==
 
== Solution ==
  
−
We need to find the least common multiple of the four numbers given. That is, the next time they will be together.
+
By translating the words in the problem into the language of mathematics, the problem is telling us to find the least common multiple of the four numbers given.  
−
First, find the least common multiple of <math> 3 </math> and <math> 4 </math>.
 
  
−
<math> 3 \times 4 = 12 </math>.
+
<math>\textrm{LCM}(3, 4, 6, 7) = \textrm{LCM}(3, 2^2, 2 \cdot 3, 7) = 2^2 \cdot 3 \cdot 7 = 84</math>
  
−
Find the least common multiple of <math> 12 </math> and <math> 6 </math>.
+
So the answer is <math>\boxed{\textbf{(B) } 84} </math>.
−
Since <math> 12 </math> is a multiple of <math> 6 </math>, the least common multiple is <math> 12 </math>.
+
 
−
Lastly, the least common multiple of <math> 12 </math> and <math> 7 </math> is <math> 12 \times\ 7 = \boxed{\textbf{(B) }84} </math>.
+
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/ts2x9Q0XVM0?si=uxpzYXU3VacX2lKm
 +
 
 +
~Thesmartgreekmathdude
 +
 
 +
== See Also ==
 +
 
 +
{{AMC10 box|year=2001|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 21:30, 13 August 2024

Problem

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?

$\textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252$

Solution

By translating the words in the problem into the language of mathematics, the problem is telling us to find the least common multiple of the four numbers given.

$\textrm{LCM}(3, 4, 6, 7) = \textrm{LCM}(3, 2^2, 2 \cdot 3, 7) = 2^2 \cdot 3 \cdot 7 = 84$

So the answer is $\boxed{\textbf{(B) } 84}$.

Video Solution by Daily Dose of Math

https://youtu.be/ts2x9Q0XVM0?si=uxpzYXU3VacX2lKm

~Thesmartgreekmathdude

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 10 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=2001_AMC_10_Problems/Problem_8&oldid=224984"