The pattern once again beings with Goose-Goose-Duck, so this goes on exactly like before:Įither $j 1\leqslant 3$ and the $n$-th students survives from others begin tagged Goose before him or else we must have $j\equiv 2 \pmod 3$ in order for $n$ to avoid exclusion. We're now reduced to $j 1$ students and the $n$-th student is now the $(j 1)$-th. Remember that to reach this point, we had to assume $k 1>3$, so necessarily $j>0$. Suppose hence that $k = 3j 2$, so we had $3j 3$ students. Hence, in order for the $(k 1)$-th student to avoid exclusion, we must have $k 1 \equiv 0 \iff$ $k \equiv2 \pmod 3$. It follows that student $i$ $(i$ ranging from $1$ to $k 1)$ will be excluded if and only if $i \equiv 1$ or $i\equiv 2\pmod 3$. If $k 1>3$, then we'll do a full pass around the circle. This means that if $k 1\leqslant 3$, the $n$-th student will survive as a result of all others being tagged Goose before him. ![]() Now, the teacher's pattern begins with Goose-Goose-Duck. In this case, the $n$-th student is now the $(k 1)$-th. Suppose hence that $k>0$, so the game will continue. In other words, for the $n$-th student to avoid exclusion, we must have $n\equiv 1 \pmod 3$.Īfter tagging each student exactly once, we're down to a circle with $k 1$ students, so if $k=0$, the game ends right after this first pass. In the first passing around the circle, student $i$ $(i$ ranging from $1$ to $n)$, will be excluded if and only if $i \equiv 0$ or $i\equiv 2\pmod 3$. I've seen the case for $n=2$ in a Numberphile video which if expressed in base $2,$ just take the first digit and put it on the bottom, also called the Josephus problem.Ĭurrently, I figured out that $3^n$ has position $1$ win every time, so I suspect the pattern still holds, but I can't prove it. Find, with proof, all values of $n$ with $n > 2$ such that if theĬircle starts with $n$ students, then the $n$ ![]() Student $7$ (duck), student $8$ (goose), student $1$ (goose), student $4$ (duck), student $7$ (goose)Īnd student $4$ is the winner. Student $2$ (goose), student $3$ (goose), student $4$ (duck), student $5$ (goose), student $6$ (goose), This remaining student is the winner.įor instance, if there are $8$ students, the game proceeds as follows: student $1$ (duck), The teacher tags students in the pattern: duck, goose, goose, duck, goose, goose, etc., andĬontinues around the circle (re-tagging some former ducks as geese) until only one student Passes each student, he taps the student on the head and declares her a ‘duck’ or a ‘goose’.Īny student named a ‘goose’ leaves the circle immediately. The game is played asįollows: All the students stand in a circle and the teacher walks around the circle. A teacher plays the game “Duck-Goose-Goose” with his class.
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