2018 AMC 8 Problems/Problem 3

Problem

Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?

$\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}\qquad$

Solution 1

The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$

The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count.

Assuming the six people start with $1$ , Arn counts $7$ so he leaves first. Then Cyd counts $14$ , as there are $7$ numbers to be counted from this point. Then Fon, Bob, and Eve, count, $17,$ $21,$ and $27$ respectively, so last one standing is Dan. Hence, the answer would be $\boxed{\textbf{(D) }\text{Dan}}$ .

Video Solution

https://youtu.be/uK4bbVpwHGc

~savannahsolver

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2018 AMC 8 Problems/Problem 3

Contents

Problem

Solution 1

Video Solution

See Also