Difference between revisions of "2013 AMC 10B Problems/Problem 13"
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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, as <math>53-45=8</math>. Since we're starting from 1 each time, 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, as <math>53-45=8</math>. Since we're starting from 1 each time, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>. | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=12|num-a=14}} | ||
Revision as of 16:02, 27 March 2013
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 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 | ||
