2003 IMO Problems/Problem 1

$S$ is the set $\{1, 2, 3, \dots ,1000000\}$. Show that for any subset $A$ of $S$ with $101$ elements we can find $100$ distinct elements $x_i$ of $S$, such that the sets $\{a + x_i \mid a \in A\}$ are all pairwise disjoint.


Consider the set $D=\{x-y \mid x,y \in A\}$. There are at most $101 \times 100 + 1 = 10101$ elements in $D$. Two sets $A + t_i$ and $A + t_j$ have nonempty intersection if and only if $t_i - t_j$ is in $D$. So we need to choose the $100$ elements in such a way that we do not use a difference from $D$. Now select these elements by induction. Choose one element arbitrarily. Assume that $k$ elements, $k \leq 99$, are already chosen. An element $x$ that is already chosen prevents us from selecting any element from the set $x + D$. Thus after $k$ elements are chosen, at most $10101k \leq 999999$ elements are forbidden. Hence we can select one more element.

See Also

2003 IMO (Problems) • Resources
Preceded by
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Problem 2
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