Subfield: Combinatorics


Question: 
A grasshopper jumps along the real axis. He starts at point 0 and makes 2009 jumps to the right with lengths $1,2, \ldots, 2009$ in an arbitrary order. Let $M$ be a set of 2008 positive integers less than $1005 \cdot 2009$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.

Solution:
We construct the set of landing points of the grasshopper.

Case 1. $M$ does not contain numbers divisible by 2009.

We fix the numbers $2009 k$ as landing points, $k=1,2, \ldots, 1005$. Consider the open intervals $I_{k}=(2009(k-1), 2009 k), k=1,2, \ldots, 1005$. We show that we can choose exactly one point outside of $M$ as a landing point in 1004 of these intervals such that all lengths from 1 to 2009 are realized. Since there remains one interval without a chosen point, the length 2009 indeed will appear. Each interval has length 2009, hence a new landing point in an interval yields with a length $d$ also the length $2009-d$. Thus it is enough to implement only the lengths from $D=\{1,2, \ldots, 1004\}$. We will do this in a greedy way. Let $n_{k}, k=1,2, \ldots, 1005$, be the number of elements of $M$ that belong to the interval $I_{k}$. We order these numbers in a decreasing way, so let $p_{1}, p_{2}, \ldots, p_{1005}$ be a permutation of $\{1,2, \ldots, 1005\}$ such that $n_{p_{1}} \geq n_{p_{2}} \geq \cdots \geq n_{p_{1005}}$. In $I_{p_{1}}$ we do not choose a landing point. Assume that landing points have already been chosen in the intervals $I_{p_{2}}, \ldots, I_{p_{m}}$ and the lengths $d_{2}, \ldots, d_{m}$ from $D$ are realized, $m=1, \ldots, 1004$. We show that there is some $d \in D \backslash\left\{d_{2}, \ldots, d_{m}\right\}$ that can be implemented with a new landing point in $I_{p_{m+1}}$. Assume the contrary. Then the $1004-(m-1)$ other lengths are obstructed by the $n_{p_{m+1}}$ points of $M$ in $I_{p_{m+1}}$. Each length $d$ can be realized by two landing points, namely $2009\left(p_{m+1}-1\right)+d$ and $2009 p_{m+1}-d$, hence

$$
n_{p_{m+1}} \geq 2(1005-m)
\tag{1}
$$

Moreover, since $|M|=2008=n_{1}+\cdots+n_{1005}$,

$$
2008 \geq n_{p_{1}}+n_{p_{2}}+\cdots+n_{p_{m+1}} \geq(m+1) n_{p_{m+1}}
\tag{2}
$$

Consequently, by (1) and (2),

$$
2008 \geq 2(m+1)(1005-m)
$$

The right hand side of the last inequality obviously attains its minimum for $m=1004$ and this minimum value is greater than 2008 , a contradiction.

Case 2. $M$ does contain a number $\mu$ divisible by 2009.

By the pigeonhole principle there exists some $r \in\{1, \ldots, 2008\}$ such that $M$ does not contain numbers with remainder $r$ modulo 2009. We fix the numbers $2009(k-1)+r$ as landing points, $k=1,2, \ldots, 1005$. Moreover, $1005 \cdot 2009$ is a landing point. Consider the open intervals



$I_{k}=(2009(k-1)+r, 2009 k+r), k=1,2, \ldots, 1004$. Analogously to Case 1 , it is enough to show that we can choose in 1003 of these intervals exactly one landing point outside of $M \backslash\{\mu\}$ such that each of the lengths of $D=\{1,2, \ldots, 1004\} \backslash\{r\}$ are implemented. Note that $r$ and $2009-r$ are realized by the first and last jump and that choosing $\mu$ would realize these two differences again. Let $n_{k}, k=1,2, \ldots, 1004$, be the number of elements of $M \backslash\{\mu\}$ that belong to the interval $I_{k}$ and $p_{1}, p_{2}, \ldots, p_{1004}$ be a permutation of $\{1,2, \ldots, 1004\}$ such that $n_{p_{1}} \geq n_{p_{2}} \geq \cdots \geq n_{p_{1004}}$. With the same reasoning as in Case 1 we can verify that a greedy choice of the landing points in $I_{p_{2}}, I_{p_{3}}, \ldots, I_{p_{1004}}$ is possible. We only have to replace (1) by

$$
n_{p_{m+1}} \geq 2(1004-m)
$$

( $D$ has one element less) and (2) by

$$
2007 \geq n_{p_{1}}+n_{p_{2}}+\cdots+n_{p_{m+1}} \geq(m+1) n_{p_{m+1}}
$$