Difference between revisions of "2013 AMC 12B Problems/Problem 7"
(→See also) |
m (→See also) |
||
| Line 12: | Line 12: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2013|ab=B|num-b=6|num-a=8}} | ||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:35, 22 November 2013
- The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.
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, as 
. Since we're starting from 1 each time, the 53rd number said will be 
.
See also
| 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. 
