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

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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.
 
<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==
 +
 +
Consider the set <math>D=\{x-y \mid x,y \in A\}</math>. There are at most <math>101 \times 100 + 1 =
 +
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
 +
<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
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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.
  
 
==See Also==
 
==See Also==
 
{{IMO box|year=2003|before=|num-a=2}}
 
{{IMO box|year=2003|before=|num-a=2}}
 
[[Category:Olympiad Combinatorics Problems]]
 
[[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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