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Difference between revisions of "1985 AIME Problems/Problem 12"

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==Solution 10 (Linear Algebra)==
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Think of the problem as a state machine with a transition matrix. State 1 is if the bug is at A, State 2 is if the bug is not at A. In vector form we can represent state 1 as <math>\begin{bmatrix} 1 \\ 0 \end{bmatrix}</math> and state 2 as <math>\begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>. Our transition matrix <math>T =\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}</math>. Thus the probability of being in each state after k moves is <math>T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}</math>. Those familiar with linear algebra will recognize the need to diagonalize the transition matrix through eigendecomposition. In short, the eigenvalues of the transition matrix are <math>-\frac{1}{3}</math> and 1, with corresponding eigenvector basis of <math>\begin{pmatrix} 1 \\ -1 \end{pmatrix},  \begin{pmatrix} \frac{1}{3} \\ 1 \end{pmatrix}</math>. Thus <math>T = Q \Lambda Q^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{1}{3}\\ -1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}</math>
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Thus <math>T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} (-\frac{1}{3})^{k} & 0\\ 0 & 1^{k} \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{pmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} & -\frac{1}{4}(-\frac{1}{3})^{k}+\frac{1}{4}\\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} &  \frac{1}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} \\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{bmatrix}</math>.
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The probability we seek is the top entry of this vector for k = 7, or <math>\frac{182}{729}</math>.
  
 
== Remark ==
 
== Remark ==

Latest revision as of 20:38, 23 July 2024

Problem

Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$.

Solution 1 (Single Variable Recursion)

For all nonnegative integers $k,$ let $P(k)$ be the probability that the bug is at vertex $A$ when it has crawled exactly $k$ meters. We wish to find $p=P(7).$

Clearly, we have $P(0)=1.$ For all $k\geq1,$ note that after $k-1$ crawls:

  1. The probability that the bug is at vertex $A$ is $P(k-1),$ and the probability that it crawls to vertex $A$ on the next move is $0.$

  2. The probability that the bug is not at vertex $A$ is $1-P(k-1),$ and the probability that it crawls to vertex $A$ on the next move is $\frac13.$

Together, the recursive formula for $P(k)$ is \[P(k) = \begin{cases} 1 & \mathrm{if} \ k=0 \\ \frac13(1-P(k-1)) & \mathrm{if} \ k\geq1 \end{cases}.\] Two solutions follow from here:

Solution 1.1 (Recursive Formula)

We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac{182}{729}. \end{alignat*} Therefore, the answer is $n=\boxed{182}.$

~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 1.2 (Explicit Formula)

Let $P(k)=Q(k)+c$ for some function $Q(k)$ and constant $c.$ For all $k\geq1,$ the recursive formula for $P(k)$ becomes \[Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.\] Solving for $Q(k),$ we get \[Q(k)=\frac13-\frac13Q(k-1)-\frac43c.\] For simplicity purposes, we set $c=\frac14,$ which gives \[Q(k)=-\frac13Q(k-1).\] Clearly, $Q(0),Q(1),Q(2),\ldots$ is a geometric sequence with the common ratio $\frac{Q(k)}{Q(k-1)}=-\frac13.$ Substituting $P(0)=1$ and $c=\frac14$ into $P(0)=Q(0)+c$ produces $Q(0)=\frac34,$ the first term of the geometric sequence.

For all nonnegative integers $k,$ the explicit formula for $Q(k)$ is \[Q(k)=\frac34\left(-\frac13\right)^k,\] and the explicit formula for $P(k)$ is \[P(k)=\frac34\left(-\frac13\right)^k+\frac14.\] Finally, the requested probability is $p=P(7)=\frac{182}{729},$ from which $n=\boxed{182}.$

~MRENTHUSIASM

Solution 2 (Multivariable Recursion by Algebra)

Denominator

There are $3^7$ ways for the bug to make $7$ independent crawls without restrictions.

Numerator

Let $V_k$ denote the number of ways for the bug to crawl exactly $k$ meters starting from vertex $V$ and ending at vertex $A,$ where $V\in\{A,B,C,D\}$ and $k$ is a positive integer. We wish to find $A_7.$

Since the bug must crawl to vertex $B,C,$ or $D$ on the first move, we have \begin{align*} A_7&=B_6+C_6+D_6 \\ &=(A_5+C_5+D_5)+(A_5+B_5+D_5)+(A_5+B_5+C_5) \\ &=A_5+2(A_5+B_5+C_5+D_5) \\ &=A_5+2S_5, \end{align*} where $S_k=A_k+B_k+C_k+D_k.$

More generally, we get \[A_{k+2}=A_k+2S_k. \qquad\qquad (\spadesuit)\] For $S_k,$ note that

  1. Base Case:

    2. \begin{align*} S_1&=A_1+B_1+C_1+D_1 \\ &=0+1+1+1 \\ &=3. \end{align*}

  2. Recursive Case:

    3. \begin{align*} S_{k+1}&=A_{k+1}+B_{k+1}+C_{k+1}+D_{k+1} \\ &=(B_k+C_k+D_k)+(A_k+C_k+D_k)+(A_k+B_k+D_k)+(A_k+B_k+C_k) \\ &=3(A_k+B_k+C_k+D_k) \\ &=3S_k. \end{align*}

Clearly, $S_1,S_2,S_3,\ldots$ is a geometric sequence with the first term $S_1=3$ and the common ratio $\frac{S_{k+1}}{S_k}=3.$ Therefore, its explicit formula is \[S_k=3^k. \qquad\qquad (\clubsuit)\] Recall that $A_1=0.$ By $(\spadesuit)$ and $(\clubsuit),$ we rewrite $A_7$ recursively: \begin{align*} A_7 &= A_5+2S_5 \\ &=\left(A_3+2S_3\right)+2S_5 \\ &=\left(\left(A_1+2S_1\right)+2S_3\right)+2S_5 \\ &=2S_1+2S_3+2S_5 \\ &=2(3)+2\left(3^3\right)+2\left(3^5\right). \end{align*} Answer

The requested probability is \[p=\frac{A_7}{3^7}=\frac{2(3)+2\left(3^3\right)+2\left(3^5\right)}{3^7}=\frac{2(1)+2\left(3^2\right)+2\left(3^4\right)}{3^6}=\frac{182}{729},\] from which $n=\boxed{182}.$

~MRENTHUSIASM

Solution 3 (Multivariable Recursion by Table)

Define notation $V_k$ as Solution 2 does.

In fact, we can generalize the following relationships for all nonnegative integers $k:$ \begin{align*} A_0&=1, \\ B_0&=0, \\ C_0&=0, \\ D_0&=0, \\ A_{k+1}&=B_k+C_k+D_k, \\ B_{k+1}&=A_k+C_k+D_k, \\ C_{k+1}&=A_k+B_k+D_k, \\ D_{k+1}&=A_k+B_k+C_k. \\ \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c} \hspace{7mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{0} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} \\ \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{A_k} &1&0&3&6&21&60&183&546 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{B_k} &0&1&2&7&20&61&182&547 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{C_k} &0&1&2&7&20&61&182&547 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{D_k} &0&1&2&7&20&61&182&547 \\ \end{array}\] Note that the paths from $V$ to $A$ and the paths from $A$ to $V$ have one-to-one correspondence. So, we must get \[A_k+B_k+C_k+D_k=3^k\] for all values of $k.$

The requested probability is \[p=\frac{A_7}{3^7}=\frac{546}{2187}=\frac{182}{729},\] from which $n=\boxed{182}.$

~MRENTHUSIASM

Solution 4 (Single Variable Version of Solution 2)

Let $a_n$ denotes the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$.

Notice that there is respectively $1$ way to arrive at $A$ for each of the different routes after the previous $n-1$ crawls, excluding the possibility that the bug ends up at $A$ after the $(n-1)$th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation \[a_n=3^{n-1}-a_{n-1}.\] Quick calculations yield \begin{align*} a_7&=3^6-a_6\\ &=3^6-\left(3^5-3^4+3^3-3^2+3-a_1\right)\\ &=546. \end{align*} Thus, $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$.

~Nafer

Solution 5 (Dynamic Programming)

Let $A(n)$ be the probability the bug lands on vertex $A$ after crawling $n$ meters, $B(n)$ be the probability the bug lands on vertex $B$ after crawling $n$ meters, and etc.

Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the bug land on each vertex after $n$ meters is $\frac13$ the sum of the probability the bug land on other vertices after crawling $n-1$ meters. So, we have \begin{align*} A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. \end{align*} It follows that $A(n) = B(n-1) = C(n-1) = D(n-1).$

We construct the following table: \[\begin{array}{c|cccc} & & & & \\ [-2ex] n & A(n) & B(n) & C(n) & D(n) \\ [1ex] \hline & & & & \\ [-1ex] 1 & 0 & \frac13 & \frac13 & \frac13 \\ & & & & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ & & & & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ & & & & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\ & & & & \\ 5 & \frac{20}{81} & \frac{61}{243} & \frac{61}{243} & \frac{61}{243} \\ & & & & \\ 6 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} \\ & & & & \\ 7 & \frac{182}{729} & \frac{547}{2187} & \frac{547}{2187} & \frac{547}{2187} \\ [1ex] \end{array}\] Therefore, the answer is $n = \boxed{182}.$

~isabelchen

Solution 6 (Generating Functions and Roots of Unity Filter / Casework)

The generating function for a problem with this general form ($4$ states, $n$ steps) is $(x+x^2+x^3)^n$, so the generating function of interest for this problem is $(x+x^2+x^3)^7$. Our goal is to find the coefficients of every $x^{4n}$ and add them up before dividing by $3^7$. Here we have two ways to proceed:

1. Roots of Unity Filter

Let $\omega = e^{i\pi / 2}$. We have that if $G(x) = (x+x^2+x^3)^7$, then \[\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.\] From here, the desired probability is $\frac{546}{2187} = \frac{182}{729}$. Therefore, the answer is $n=\boxed{182}$.

~RedFlame2112

2. Casework

We can factor $(x+x^2+x^3)^7$ as $x^7(1+x+x^2)^7.$ The $x^{4n}$ coefficients of $(x+x^2+x^3)^7$ will be the same as the $x^{4n+1}$ coefficients of $(1+x+x^2)^7.$ The possible values for $4n+1$ would then be $1,$ $5,$ $9,$ and $13.$ Because $1+13=5+9=14,$ the coefficients of $x^1$ and $x^{13}$ are equal and so are the coefficients of $x^5$ and $x^9.$ To make an $x$ term, we need $6$ "$1$" terms and one "$x$" term multiplied together. There are $7$ ways to arrange these numbers. The coefficient of the $x^5$ term will be the sum of the number of the possible arrangements for these numbers: \begin{align*} 0000122&=105\text{ arrangements}, \\ 0001112&=140\text{ arrangements}, \\ 0011111&=21\text{ arrangements}. \end{align*} Thus, the coefficient of the $x^5$ term is $105+140+21=266.$ From here, the desired probability is $\frac{2(7+266)}{2187}=\frac{546}{2187}=\frac{182}{729}.$ Thus, our answer is $\boxed{182}.$

~BS2012

Solution 7 (Partitions)

We can find the number of different times the bug reaches vertex $A$ before the $7$th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A.$

Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obviously $f(1) = 0.$ In general for $f(x),$ the bug has three initial edges to pick from. From there, since the bug cannot return to $A$ by definition, the bug has exactly two choices. This continues from the $2$nd move up to the $(x-1)$th move. The last move must be a return to $A,$ so this move is determined. So $f(x) = 2^{x-2}3.$

Now we need to find the number of cycles by which the bug can reach $A$ at the end. Since $f(1) = 0,$ we know that $f(6)$ cannot be used, as on the $7$th move the bug cannot move from $A$ to $A.$ So we need to find the number of partitions of $7$ using only $2,3,4,5,$ and $7.$ These are $f(2)f(2)f(3),f(2)f(5),f(3)f(4),$ and $f(7).$ We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition: \[{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7) = 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53) = 546.\] Finally, this is a probability question, so we divide by $3^7,$ or $\frac{546}{3^7} = \frac{182}{3^6}.$ The answer is $n=\boxed{182}.$

Solution 8 (Sequences)

Note that this problem is basically equivalent to the following: How many distinct sequences of $8$ integers $a_1, a_2, a_3, \ldots, a_8$ are there such that $a_1 = a_8 = 1,$ $a_i \in \{1, 2, 3, 4\}$ for all $2 \leq i\leq 8,$ and $a_i \neq a_{i+1}$ for all $1 \leq i \leq 7$?

Now consider the $8$ integers modulo $4.$ Let $b_1, b_2, b_3, \ldots, b_7$ be a new sequence of integers such that $b_i = a_{i+1} - a_i \mod 4$ for all $1 \leq i \leq 7.$

Note that the condition is equivalent to that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7,$ and since $a_1 \mod 4 = a_8 \mod 4,$ $b_1 + b_2 + \cdots + b_7$ must be a multiple of $4.$

Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers $(b_1, b_2, \ldots, b_7)$ such that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7$ and $b_1 + b_2 + \cdots + b_7$ is congruent to $0 \mod 4.$

We will now proceed with casework on the number of $2$s in the septuple.

One $2$: There are $\dbinom{7}{1} = 7$ ways to arrange the $2$, and $\dbinom{6}{0} + \dbinom{6}{2} + \dbinom{6}{4} + \dbinom{6}{6} = 2^5$ valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the $1$s and $3$s.

Three $2$s: There are $\dbinom{7}{3} = 35$ ways to arrange the $2$s, and $\dbinom{4}{1} + \dbinom{4}{3} = 2^3$ valid ways to arrange the $1$s and $3$s.

Five $2$s: There are $\dbinom{7}{5} = 21$ ways to arrange the $2$s, and $\dbinom{2}{0} + \dbinom{2}{2} = 2$ valid ways to arrange the $1$s and $3$s.

Adding up our cases, we obtain $7 \cdot 32 + 35 \cdot 8 + 21 \cdot 2 = 546$ valid sequences. Dividing by the total number of paths without restrictions, $3^7 = 2187,$ our desired probability is $\frac{546}{2187} = \frac{182}{729},$ giving an answer of $\boxed{182}.$

~fidgetboss_4000

Solution 9

We instead find the probability that the bug is NOT at vertex $A$ after crawling $n$ meters (equivalent to moving $n$ times). Call $A_n$ the probability that the bug IS at vertex $A$ after $n$ moves; call $O_n$ the probability that the bug is on some other vertex. We have the following recurrence relations. \[A_n = \frac{1}{3}O_{n-1}\] \[O_n = A_{n-1} + \frac{2}{3}O_{n-1}\] However, we can calculate $A_{n-1}$ in terms of $O_n$; take $n = k-1$, and we have \[A_{k-1} = \frac{1}{3}O_{k-2}\]. Substituting this into our equation for $O$, we have that \[O_n = \frac{1}{3}O_{n-2} + \frac{2}{3}O_{n-1}\]. We also have the conditions that $O_0 = 0$ (the bug will not be at vertex other than $A$ on the "0th" move) and $O_1 = 1$ (the bug will be at a vertex other than $A$ after the $1st$ move). Iteratively calculating $O_7$, we find that it is equal to $\frac{547}{729}$ (I will not do this calculation here; you can do it manually if you wish to check). However, this is the probability that the ant is NOT at vertex $A$ after $7$ moves; then its complement, $\frac{182}{729}$ is the probability that the ant IS at vertex $A$ after $7$ moves, so our answer is $\boxed{182}$.

~ cxsmi

Solution 10 (Linear Algebra)

Think of the problem as a state machine with a transition matrix. State 1 is if the bug is at A, State 2 is if the bug is not at A. In vector form we can represent state 1 as $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and state 2 as $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Our transition matrix $T =\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}$. Thus the probability of being in each state after k moves is $T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Those familiar with linear algebra will recognize the need to diagonalize the transition matrix through eigendecomposition. In short, the eigenvalues of the transition matrix are $-\frac{1}{3}$ and 1, with corresponding eigenvector basis of $\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} \frac{1}{3} \\ 1 \end{pmatrix}$. Thus $T = Q \Lambda Q^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{1}{3}\\ -1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}$

Thus $T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} (-\frac{1}{3})^{k} & 0\\ 0 & 1^{k} \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{pmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} & -\frac{1}{4}(-\frac{1}{3})^{k}+\frac{1}{4}\\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} & \frac{1}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} \\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{bmatrix}$.

The probability we seek is the top entry of this vector for k = 7, or $\frac{182}{729}$.

Remark

Here is a similar problem from another AIME test: 2003 AIME II Problem 13, in which we have an equilateral triangle instead.

There is another similar problem from the AMC8: 2022 AMC 8 Problems/Problem 25, where we have the same question, just with less steps

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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