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

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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
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==Solution==
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Looking at the values, we notice that <math>11-7=4</math>, <math>9-5=4</math> and <math>6-2=4</math>. This means we are looking for a value that is four less than a mulitple of 11, 9 and 6. The least common multiple of these number is <math>11*3^2*2=198</math>, so the numbers that fulfill this can be written as <math>198k-4</math>, where k is a positive integer.  This value is only a three digit integer when <math>k</math> is <math>1, 2, 3, 4</math> or <math>5</math>, which output <math>194, 392, 590, 788,</math> and <math>986</math> respectively. Thus we have 5 values, so our answer is <math>\boxed{E}</math>
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{{AMC8 box|year=2018|num-b=20|num-a=22}}
 
{{AMC8 box|year=2018|num-b=20|num-a=22}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:46, 21 November 2018

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

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 mulitple of 11, 9 and 6. The least common multiple of these number is $11*3^2*2=198$, 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 output $194, 392, 590, 788,$ and $986$ respectively. Thus we have 5 values, so our answer is $\boxed{E}$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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