Difference between revisions of "2018 AMC 8 Problems/Problem 3"

(Problem 3)
(Undo revision 110681 by Solvingking (talk))
(Tag: Undo)
Line 1: Line 1:
 
==Problem 3==
 
==Problem 3==
What is the sum of the first 3 numbers
+
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?
  
 
<math>\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}</math>
 
<math>\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}</math>

Revision as of 21:59, 29 October 2019

Problem 3

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?

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

Solution

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 $\boxed{\text{(D) Dan}}$.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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