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Difference between revisions of "2017 AMC 8 Problems/Problem 12"
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<math>\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</math>
 
<math>\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</math>
  
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==Solution==
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==Solution 1==
  
 
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>
 
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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==Solution 2==
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Call the number we want to find <math>n</math>. We can say that
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<cmath>n \equiv 1 \mod 4</cmath>
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<cmath>n \equiv 1 \mod 5</cmath>
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<cmath>n \equiv 1 \mod 6.</cmath>
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We can also say that <math>n-1</math> is divisible by <math>4,5</math> and <math>6.</math>
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Therefore, <math>n-1=lcm(4,5,6)=69</math>, so <math>n=60+1=61</math> which is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:36, 31 December 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 1

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 $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

Solution 2

Call the number we want to find $n$. We can say that \[n \equiv 1 \mod 4\] \[n \equiv 1 \mod 5\] \[n \equiv 1 \mod 6.\] We can also say that $n-1$ is divisible by $4,5$ and $6.$ Therefore, $n-1=lcm(4,5,6)=69$, so $n=60+1=61$ which is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

Video Solution

https://youtu.be/SwgXyYFIuxM

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
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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