Difference between revisions of "2001 AMC 10 Problems/Problem 8"

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== Solution ==
 
== Solution ==
  
We need to find the least common multiple of the three numbers given.  
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We need to find the least common multiple of the four numbers given.  
  
<math>\textrm{LCM}(15, 20, 25) = \textrm{LCM}(3 \cdot 5, 2^2 \cdot 5, 5^2) = 2^2 \cdot 3 \cdot 5^2 = 300</math>
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<math>\textrm{LCM}(3, 4, 6, 7) = \textrm{LCM}(3, 2^2, 2 \cdot 3, 7) = 2^2 \cdot 3 \cdot 7 = 84</math>
  
300 minutes equals 5 hours. So the bell will ring 5 hours past 12:00, which is 17:00.
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So the answer is <math>\boxed{\textbf{(B) } 84} </math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 14:31, 21 July 2023

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

We need 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 $\boxed{\textbf{(B) } 84}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
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

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