Difference between revisions of "2018 AMC 8 Problems/Problem 3"
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Assuming the six 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,</math> <math>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 six 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,</math> <math>21,</math> and <math>27</math> respectively, so last one standing is Dan. Hence, the answer would be <math>\boxed{\text{(D) Dan}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/uK4bbVpwHGc | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 10:10, 18 February 2022
Contents
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?
Solution 1
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 six 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 
.
Video Solution
~savannahsolver
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. 
