Difference between revisions of "2003 AIME II Problems/Problem 13"

Revision as of 19:00, 15 July 2011

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 $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution

Solution 1

After n moves, there are $2^n$ possible paths for the bug. 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 $n-1$ moves. So after 1 move, there are 0 ways the get to the start. After 2 moves, there are $2^1-0$ ways to get to the start. After 3 moves, there are $2^2-(2^1-0)$ ways to get to the start. Thus after 10 moves, there are $2^9-(2^8-(2^7-(2^6-(2^5-(2^4-(2^3-(2^2-2)))))))= 342$ ways to get to the start, so the probability is $342/1024=171/512$ and the answer is $683$ .

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., $\#CW - \#CCW \equiv 0 \pmod{3}$ . Since $\#CW + \#CCW = 10$ , it is only possible that $(\#CW,\, \#CCW) = (5,5), (8,2), (2,8)$ .

In the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \choose 5}$ paths. In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise. Hence the total number of ways to return to the original vertex is ${10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342$ . Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\frac{342}{2^{10}} = \frac{171}{512}$ , and the answer is $683$ .

@@ Line 5: / Line 5: @@
 ===Solution 1===
-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>.
+After n moves, there are <math>2^n</math> possible paths for the bug. 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===

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

Page

Toolbox

Search