Difference between revisions of "2018 AMC 8 Problems/Problem 3"
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The five numbers which cause people to leave the circle are <math>7, 14, 17, 21,</math> and <math>27.</math> | The five numbers which cause people to leave the circle are <math>7, 14, 17, 21,</math> and <math>27.</math> | ||
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+ | 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 five people start with <math>1</math>, Arn counts <math>7</math> so he leaves first. Then Cyd counts <math>14</math>, as there are <math>7</math> numbers to be counted from this point. Then Fon, Bob, and Eve, count <math>17, 21,</math> and <math>27</math> respectively, so last one standing is Dan. Hence the answer would be <math>\boxed{\text{(D) Dan}}</math> | Assuming the five people start with <math>1</math>, Arn counts <math>7</math> so he leaves first. Then Cyd counts <math>14</math>, as there are <math>7</math> numbers to be counted from this point. Then Fon, Bob, and Eve, count <math>17, 21,</math> and <math>27</math> respectively, so last one standing is Dan. Hence the answer would be <math>\boxed{\text{(D) Dan}}</math> |
Revision as of 01:07, 26 November 2018
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?
Solution
The five numbers which cause people to leave the circle are and
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 five people start with , Arn counts
so he leaves first. Then Cyd counts
, as there are
numbers to be counted from this point. Then Fon, Bob, and Eve, count
and
respectively, so last one standing is Dan. Hence the answer would be
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
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.