Difference between revisions of "2017 AMC 8 Problems/Problem 12"

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==Solution==
 
==Solution==
  
The <math>\text{lcm}(4,5,6)</math> is 60. Since <math>60+1=61</math>, and that is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math>
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The <math>\operatorname{LCM}(4,5,6)</math> is <math>60</math>. Since <math>60+1=61</math>, and that is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 23:26, 4 January 2018

Problem 12

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

$\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124$

Solution

The $\operatorname{LCM}(4,5,6)$ is $60$. Since $60+1=61$, and that is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

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