Difference between revisions of "2003 AIME II Problems/Problem 13"
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===Solution 1=== | ===Solution 1=== | ||
− | After n moves, there are <math>2^n</math> possible paths for the | + | After n moves, there are <math>2^n</math> possible paths for the ant. But the key is to realize that the number of paths that get back the the start after n moves is the same as the number of paths the do NOT get the ant to the start after <math>n-1</math> moves. So after 1 move, there are 0 ways the get to the start. After 2 moves, there are <math>2^1-0</math> ways to get to the start. After 3 moves, there are <math>2^2-(2^1-0)</math> ways to get to the start. Thus after 10 moves, there are <math>2^9-(2^8-(2^7-(2^6-(2^5-(2^4-(2^3-(2^2-2)))))))= 342</math> ways to get to the start, so the probability is <math>342/1024=171/512</math> and the answer is <math>683</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 18:52, 18 July 2011
Contents
[hide]Problem
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is where and are relatively prime positive integers, find
Solution
Solution 1
After n moves, there are possible paths for the ant. But the key is to realize that the number of paths that get back the the start after n moves is the same as the number of paths the do NOT get the ant to the start after moves. So after 1 move, there are 0 ways the get to the start. After 2 moves, there are ways to get to the start. After 3 moves, there are ways to get to the start. Thus after 10 moves, there are ways to get to the start, so the probability is and the answer is .
Solution 2
Consider there to be a clockwise and a counterclockwise direction around the triangle. Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., . Since , it is only possible that .
In the first case, we pick out of the ant's steps to be clockwise, for a total of paths. In the second case, we choose of the steps to be clockwise, and in the third case we choose to be clockwise. Hence the total number of ways to return to the original vertex is . Since the ant has two possible steps at each point, there are routes the ant can take, and the probability we seek is , and the answer is .
See also
2003 AIME II (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 | ||
All AIME Problems and Solutions |