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

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==Problem 12==
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==Problem==
 
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?
 
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?
  
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==Solution==
 
==Solution==
  
Finding the LCM of <math>4</math>, <math>5</math>, and <math>6</math>; we find it is <math>60</math>. Now, <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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Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. 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>
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==Video Solution==
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https://youtu.be/SwgXyYFIuxM
  
 
==See Also==
 
==See Also==
{{AMC8 box|year=2017|num-b=11|after=13}}
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{{AMC8 box|year=2017|num-b=11|num-a=13}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:02, 18 January 2021

Problem

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

Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The $\operatorname{LCM}(4,5,6)$ is $60$. Since $60+1=61$, and that is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

Video Solution

https://youtu.be/SwgXyYFIuxM

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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