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Difference between revisions of "2003 IMO Problems/Problem 1"

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<math>S</math> is the set <math>\set{1, 2, 3, \dots ,1000000}</math>. Show that for any subset <math>A</math> of <math>S</math> with <math>101</math> elements we can find <math>100</math> distinct elements <math>x_i</math> of <math>S</math>, such that the sets <math>\set{a + x_i \mid a \in A}</math> are all pairwise disjoint.
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<math>S</math> is the set <math>\{1, 2, 3, \dots ,1000000\}</math>. Show that for any subset <math>A</math> of <math>S</math> with <math>101</math> elements we can find <math>100</math> distinct elements <math>x_i</math> of <math>S</math>, such that the sets <math>\{a + x_i \mid a \in A\}</math> are all pairwise disjoint.
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==Solution==
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Consider the set <math>D=\{x-y \mid x,y \in A\}</math>. There are at most <math>101 \times 100 + 1 =
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10101</math> elements in <math>D</math>. Two sets <math>A + t_i</math> and <math>A + t_j</math> have nonempty intersection if and only if
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<math>t_i - t_j</math>
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is in <math>D</math>. So we need to choose the <math>100</math> elements in such a way that we do not use a
 +
difference from <math>D</math>.
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Now select these elements by induction. Choose one element arbitrarily. Assume that
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<math>k</math> elements, <math>k \leq 99</math>, are already chosen. An element <math>x</math> that is already chosen prevents us
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from selecting any element from the set <math>x + D</math>. Thus after <math>k</math> elements are chosen, at most
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<math>10101k \leq 999999</math> elements are forbidden. Hence we can select one more element.
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==See Also==
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{{IMO box|year=2003|before=|num-a=2}}
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[[Category:Olympiad Combinatorics Problems]]

Latest revision as of 15:09, 24 November 2019

$S$ 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.

Solution

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
' 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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