Difference between revisions of "2018 AMC 8 Problems/Problem 21"

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==Video Solution==
 
==Video Solution==
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https://youtu.be/CPQpkpnEuIc - Happytwin
  
 
https://youtu.be/PTwMDbsz2xI
 
https://youtu.be/PTwMDbsz2xI
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https://youtu.be/7an5wU9Q5hk?t=939
  
 
==See Also==
 
==See Also==

Revision as of 19:39, 16 January 2021

Problem 21

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution 1

Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three digit integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$

Video Solution

https://youtu.be/CPQpkpnEuIc - Happytwin

https://youtu.be/PTwMDbsz2xI

https://youtu.be/7an5wU9Q5hk?t=939

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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