Difference between revisions of "2013 AMC 10B Problems/Problem 4"
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Notice that for a number <math>n \le 201</math>, the place in which it is counted is the same as <math>201 - n + 1 = 202 - n</math>. (This is evident due to the fact that 201 is the 1st number counted, and every number after that increments the place by one). Thus, 53 is the <math>202 - 53 = \boxed{\textbf{(D) }149}</math>th number counted. | Notice that for a number <math>n \le 201</math>, the place in which it is counted is the same as <math>201 - n + 1 = 202 - n</math>. (This is evident due to the fact that 201 is the 1st number counted, and every number after that increments the place by one). Thus, 53 is the <math>202 - 53 = \boxed{\textbf{(D) }149}</math>th number counted. | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=3|num-a=5}} | ||
Revision as of 15:59, 27 March 2013
Problem
When counting from 
to 
, 
is the 
number counted. When counting backwards from to , is the 
number counted. What is 
?
Solution
Notice that for a number 
, the place in which it is counted is the same as 
. (This is evident due to the fact that 201 is the 1st number counted, and every number after that increments the place by one). Thus, 53 is the 
th number counted.
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
