2018 AIME I Problems/Problem 10

Problem

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$

$[asy] unitsize(32); draw(unitcircle); draw(scale(2) * unitcircle); for(int d = 90; d < 360 + 90; d += 72){ draw(2 * dir(d) -- dir(d)); } real s = 4; dot(1 * dir( 90), linewidth(s)); dot(1 * dir(162), linewidth(s)); dot(1 * dir(234), linewidth(s)); dot(1 * dir(306), linewidth(s)); dot(1 * dir(378), linewidth(s)); dot(2 * dir(378), linewidth(s)); dot(2 * dir(306), linewidth(s)); dot(2 * dir(234), linewidth(s)); dot(2 * dir(162), linewidth(s)); dot(2 * dir( 90), linewidth(s)); defaultpen(fontsize(10pt)); real r = 0.05; label("$A$", (1-r) * dir( 90), -dir( 90)); label("$B$", (1-r) * dir(162), -dir(162)); label("$C$", (1-r) * dir(234), -dir(234)); label("$D$", (1-r) * dir(306), -dir(306)); label("$E$", (1-r) * dir(378), -dir(378)); label("$F$", (2+r) * dir(378), dir(378)); label("$G$", (2+r) * dir(306), dir(306)); label("$H$", (2+r) * dir(234), dir(234)); label("$I$", (2+r) * dir(162), dir(162)); label("$J$", (2+r) * dir( 90), dir( 90)); [/asy]$

Solution 1

We divide this up into casework. The "directions" the bug can go are $\text{Clockwise}$ , $\text{Counter-Clockwise}$ , and $\text{Switching}$ . Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles. An example string of length fifteen that gets the bug back to $A$ would be $ISSIIISOOSISSII$ . For the bug to end up back at $A$ , the difference between the number of $I$ 's and $O$ 's must be a multiple of $5$ .

Case 1 -- There are 15 more $I$ 's than $O$ 's.: There is clearly $1$ way for this to happen.

Case 2 -- There are $5$ more $I$ 's than $O$ 's.

We split this case up into several sub-cases based on the number of $S$ 's.

Sub-case 1 -- There are $10$ $S$ 's and $5$ $I$ 's.: Notice that the number of ways to order the $I$ 's and $O$ 's are independent assortments because the $I$ 's must be in the "even" spaces between $S$ 's (i.e. before the 1st $S$ , between the 2nd and 3rd $S$ 's, etc.), while the $O$ 's must be in the "odd" spaces.

There are $6$ places to put the $I$ 's (after the 0th, 2nd, 4th, 6th, 8th, and 10th $S$ 's), and $4$ places to put the (0) $O$ 's. We use stars and bars to get an answer of $\binom{10}{5}\binom{4}{0}$
Sub-case 2 -- There are $8$ $S$ 's, $6$ $I$ 's, and $1$ $O$ .: Similarly and by using stars and bars, we get an amount of $\binom{10}{4}\binom{4}{1}$

All the other sub-cases are similar, with a total of $\binom{10}{5}\binom{4}{0}+\binom{10}{4}\binom{4}{1}+\cdots+\binom{10}{1}\binom{4}{4}=\binom{14}{5}=2002$ by Vandermonde's Identity.

Case 3 -- There are $5$ more $O$ 's than $I$ 's.

This case is similar to the other case.

Here is an example of a sub-case for this case.

Sub-case: There are $10$ $S$ 's and $5$ $O$ 's.; There are $\binom{9}{4}\binom{5}{0}$ ways to do this.

We can see now that the pattern is going to be $\binom{9}{4}\binom{5}{0}+\binom{9}{3}\binom{5}{1}+\cdots+\binom{9}{0}\binom{5}{4}=\binom{14}{4}=1001$ .

So, the total number of ways is $1+2002+1001=3004$ which gives $\boxed{004}$ as the answer.

Solution 2 (Official MAA)

Note that the set of sequences of moves the bug makes is in bijective correspondence with the set of strings of $X$ s and $Y$ s of length $15$ , where $X$ denotes a move which is either counterclockwise or inward along a spoke and $Y$ denotes a move which is either clockwise or outward along a spoke. (The proof of this basically boils down to the fact that which one depends on whether the bug is on the inner wheel or the outer wheel.) Now the condition that the bug ends at A implies that the difference between the number of $X$ s and the number of $Y$ s is a multiple of $5$ , and so we must have either $4$ , $9$ , or $14$ $X$ s within the first fourteen moves with the last move being an $X$ . This implies the answer is $\binom{14}4+\binom{14}9+\binom{14}{14} = 3004\equiv \boxed{004}\pmod{1000}.$

Solution 3 (Similar To Solution 2 But Modified To Clarify)

Let an $O$ signal a move that ends in the outer circle and $I$ signal a move that ends in the inner circle. Now notice that for a string of $15$ moves to end at $A$ , the difference between $O$ 's and $I$ 's in the string must be a multiple of $5$ .

$15$ $I$ 's: Trivially $1$ case.

$5$ $O$ 's and $10$ $I$ 's: Since the string has to end in an $I$ for the bug to land on $A$ , there are a total of $\binom{14}{5}=2002$ ways to put $5$ $O$ 's in the first $14$ moves.

$10$ $O$ 's and $5$ $I$ 's: Similarly there are $\binom{14}{4}=1001$ ways to put $5-1=4$ $I$ 's in the first $14$ moves.

$15$ $O$ 's: Impossible since the string has to end with an I.

This brings us an answer of $1+2002+1001=3004 \equiv \boxed{004} \pmod{1000}$ .

-Solution by mathleticguyyyyyyyyy- ~Edited by ike.chen

Solution 4 (Recursion)

Define $A_n$ to be the number of sequences of length $n$ that ends at $A$ and similarly for the other spokes. Also let $S_n=A_n+B_n+C_n+D_n+E_n+F_n+G_n+H_n+I_n+J_n$ Apparently everytime the bug has $2$ choices for its next move, thus we have $S_n=2^n$ . Now we attempt to find a recursive formula for $A_n$ . $\begin{align*} A_n&=J_{n-1}+E_{n-1} \\ &=(I_{n-2}+A_{n-2})+(D_{n-2}+F_{n-2}) \\ &=(H_{n-3}+B_{n-3})+(J_{n-3}+E_{n-3})+(G_{n-3}+C_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(B_{n-3}+C_{n-3}+E_{n-3}+G_{n-3}+H_{n-3}+J_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(S_{n-3}-(I_{n-3}+A_{n-3}+D_{n-3}+F_{n-3}))+A_{n-2} \\ &=2^{n-3}-(J_{n-2}+E_{n-2})+A_{n-2} \\ &=2^{n-3}-A_{n-1}+A_{n-2} \\ \end{align*}$ Computing a few easy terms we have $A_0=0$ , $A_1=0$ , $A_2=1$ , $A_3=0$ , $A_4=3$ . Continuing the process yields $A_{15}=3\boxed{004}$ .

~ Nafer

Solution 5 (Stars and Bars)

We approach this problem by casework for how many times the ant passes on a pair of spokes in and out (switching back and forth between the inner and outer circle). Since each point on the outer circle is connected to a point in the inner circle by a spoke, the points have a one-to-one correspondence. Therefore, we can rethink the scenario by switching directions back and forth in the inner circle (whenever we "go on a spoke" we change directions). The number of "moves" the ant makes in this inner circle would then be 15 minus the (number of times we switch directions) since each time we switch directions, it would act as going on a spoke in our original scenario. Also, note that we will have to end up at point A in the inner circle, so there will be an even number of "switches" that the ant moves from the outer and inner circles so we can think of it in pairs. We proceed as follows:

Case 1: 0 "pairs" The ant would have to only be in the inner circle and go in one direction. Hence there is $1$ way.

Case 2: 1 "pair" That means we can think about $13$ "moves" in the inner circle where we switch directions $2$ times (counterclockwise-clockwise-counterclockwise). Note that the number of counterclockwise and clockwise moves are equivalent modulo $5$ , which is the only way they can "cancel out". Let the number of counterclockwise moves be $x$ and clockwise moves be $y$ . Then, the only two feasible solutions for $(x,y)$ are $(4,9)$ and $(9,4)$ . Note that $x$ and $y$ can be $0$ , which is equivalent to switching and immediately switching back. Therefore, we use stars and bars to obtain $10+5=15$ ways for this case.

Case 3: 2 "pairs" Using similar logic, we obtain $(3,8)$ and $(8,3)$ . Using stars and bars once again, we obtain $90+180=270$ ways for this case.

Case 4: 3 "pairs" Again, the only solutions for $(x,y)$ are $(2,7)$ and $(7,2)$ . Using stars and bars, we obtain $360+730=1080$ .

Case 5: 4 "pairs" The only solutions are $(1,6)$ and $(6,1)$ . Using stars and bars, we obtain $840+420=1260$ .

Case 6: 5 "pairs" The only solutions are $(0,5)$ and $(5,0)$ . Using stars and bars, we obtain $252+126=378$ . Note that there are no more cases since $x$ and $y$ have to be equivalent modulo 5, which would need to make $x$ or $y$ negative, hence not possible. Adding all the cases, we get $1+15+270+1080+1260+378=3004$ . Taking modulo 1000, we obtain $3004 \equiv \boxed{004} \pmod{1000}$

The process of using stars and bars was the number of ways to group the counterclockwise and clockwise moves. For instance, if there are n "pairs" of switches, then based on the solutions for $(x,y)$ , we can partition $x$ into $n+1$ subsets (indistinguishable) and $y$ into $n$ subsets respectively. Then, we multiply both results. This is equivalent to putting $x$ objects into $n+1$ bins where a bin can have 0 objects since the subsets are indistinguishable. We use the formula $\dbinom{a+b-1}{a-1}$ for putting $b$ objects into $a$ bins. Plugging in $x$ and $n+1$ for $a$ and $b$ respectively helps us reach the conclusion for each case.

Also, we derived the solutions $(x,y)$ since they have to differ by a multiple of $5$ . The only feasible difference is $5$ . Therefore, $2x+5$ is equivalent to the number of moves in the inner circle in our scenario. We can also switch $(x,y)$ to $(y,x)$ giving us two feasible solutions for each case.

-Magnetoninja

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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