A certain frog that was placed on a vertex of a convex polygon chose to jump to another vertex, either clockwise skipping one vertex, either counterclockwise skipping two vertexes, and repeated the procedure. If the number of jumps that the frog made is equal to the number of sides of the polygon, the frog has passed through all its vertexes and ended up on the initial vertex, what´s the set formed by all the possible values that this number can take? Andrei Eckstein
Problem
Source: Stars of Mathematics 2017, Juniors, Problem 3
Tags: discrete maths, combinatorics
28.02.2019 05:28
The polygon has $n$ vertices. If $n$ is odd or $3$ is not a divisor of $n$ it is easy to see that the frog can jump $n$ times, satisfiying the condition. Suppose $6|n$. Then the frog cannot jump just by jumping clockwise $n$ times or jumping counterclockwise $n$ times. Suppose the frog jumped clockwise $a$ times and counterclockwise $b$ times. Then $a+b=n$. Also, the frog has to end up on the initial vertex, so $n|\left(2a-3b\right)$. The possible values are $2a-3b=-2n,-n,0,n$. In each case, $a=\frac{n}{5}$, $a=\frac{2n}{5}$, $a=\frac{2n}{5}$, $a=\frac{2n}{5}$. So $n$ has to be a multiple of $5$. Therefore $30|n$. Also, if $30|n$, the frog can move by $1\rightarrow 3\rightarrow 5 \rightarrow 2 \rightarrow 4 \rightarrow 6 \rightarrow \cdots \rightarrow \left(n-1\right) \rightarrow 1$. Therefore the set of the possible values is the set of natural numbers which are not a multiple of $6$ or which are a multiple of $30$.
06.03.2019 20:05
why is 2a-3b = 2n why not anything greater than 2 can frog complete the hops if it goes forward and backward in 2n hops Did not understand the logic.