Difference between revisions of "2013 AMC 12B Problems/Problem 7"
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We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>. Since we are starting from 1 every turn, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>. | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>. Since we are starting from 1 every turn, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/a-3CAo4CoWc | ||
| + | |||
| + | ~[[User:icematrix2|icematrix2]] | ||
== See also == | == See also == | ||
Revision as of 21:08, 30 September 2020
- The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.
Contents
Problem
Jo and Blair take turns counting from 
to one more than the last number said by the other person. Jo starts by saying 
, so Blair follows by saying 
. Jo then says 
, and so on. What is the 
number said?
Solution
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", 
numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because 
. Since we are starting from 1 every turn, the 53rd number said will be 
.
Video Solution
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AMC 10 Problems and Solutions | ||
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
