Difference between revisions of "2021 AIME II Problems/Problem 8"

(Solution)
(Solution)
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==Solution==
 
==Solution==
 
The way we approach the problem is, we need a chart for the possible outcomes of the ant. We are trying to find the probability the ant is on the top of the cube, so let's make the chart. Let BO be "Base but originally there", BG "Base but just got there, NO, and NG. The  first choice is easy to compute. There is one choice to go on top, and 2 choices to go on the bottom.  
 
The way we approach the problem is, we need a chart for the possible outcomes of the ant. We are trying to find the probability the ant is on the top of the cube, so let's make the chart. Let BO be "Base but originally there", BG "Base but just got there, NO, and NG. The  first choice is easy to compute. There is one choice to go on top, and 2 choices to go on the bottom.  
 +
 
<math>\begin{array}{r|c|c|c|c|c|c|c|c}   
 
<math>\begin{array}{r|c|c|c|c|c|c|c|c}   
 
\text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   
 
\text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   
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\text{TG} &0&&&&&&&   
 
\text{TG} &0&&&&&&&   
 
\end{array}</math>
 
\end{array}</math>
 +
 
We can keep doing this process, until the 8th step.  
 
We can keep doing this process, until the 8th step.  
 +
 
<math>\begin{array}{r|r|r|r|r|r|r|r|r}   
 
<math>\begin{array}{r|r|r|r|r|r|r|r|r}   
 
\text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   
 
\text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   
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\text{TO} &0&2&6&10&14&26&62&138   
 
\text{TO} &0&2&6&10&14&26&62&138   
 
\end{array}</math>
 
\end{array}</math>
 +
 
Since we want the fraction when the ant is on top, <math>\dfrac{204}{384}</math> = <math>\dfrac{17}{32}</math>. Our answer is 17 + 32 = <math>\boxed{049}</math>
 
Since we want the fraction when the ant is on top, <math>\dfrac{204}{384}</math> = <math>\dfrac{17}{32}</math>. Our answer is 17 + 32 = <math>\boxed{049}</math>
  

Revision as of 11:30, 23 March 2021

Problem

An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly 8 moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

The way we approach the problem is, we need a chart for the possible outcomes of the ant. We are trying to find the probability the ant is on the top of the cube, so let's make the chart. Let BO be "Base but originally there", BG "Base but just got there, NO, and NG. The first choice is easy to compute. There is one choice to go on top, and 2 choices to go on the bottom.

$\begin{array}{r|c|c|c|c|c|c|c|c}    \text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   \text{BG} &0&&&&&&& \\ \hline   \text{BO} &2&&&&&&& \\ \hline   \text{TO} &1&&&&&&& \\ \hline   \text{TG} &0&&&&&&&    \end{array}$

We can keep doing this process, until the 8th step.

$\begin{array}{r|r|r|r|r|r|r|r|r}    \text{Steps} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline   \text{BG} &0&0&2&6&10&14&26&62 \\ \hline   \text{BO} &2&2&2&6&18&38&66&118 \\ \hline   \text{TG} &1&2&2&2&6&18&38&66 \\ \hline   \text{TO} &0&2&6&10&14&26&62&138    \end{array}$

Since we want the fraction when the ant is on top, $\dfrac{204}{384}$ = $\dfrac{17}{32}$. Our answer is 17 + 32 = $\boxed{049}$

~Arcticturn

See also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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