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Difference between revisions of "2002 AMC 10P Problems/Problem 15"

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== Solution ==
 
== Solution ==
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There are twelve pairs <math>\{ 1, 9 \}</math>, <math>\{ 2, 10 \}</math>, <math>\{ 3, 11 \}</math>, <math>\{ 4, 11 \}</math>, \; <math>\dots \; \{ 12, 20 \}</math> in <math>\{ 1, 2, 3, \; \dots \; , 20 \}</math> that differ by 8. If we take <math>n = 12</math>, it could be that we selected one element from each pair for the subset: the condition may not be fulfilled. the In order to select at least one pair, it is necessary to select <math>\textbf{(D)} \; 13</math> elements (Pigeonhole principle).
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There are twelve pairs <math>\{ 1, 9 \}</math>, <math>\{ 2, 10 \}</math>, <math>\{ 3, 11 \}</math>, <math>\{ 4, 11 \}</math>, <math>\; \dots \; \{ 12, 20 \}</math> in <math>\{ 1, 2, 3, \; \dots \; , 20 \}</math> that differ by 8. If we take <math>n = 12</math>, it could be that we selected one element from each pair for the subset: the condition may not be fulfilled. the In order to select at least one pair, it is necessary to select <math>\textbf{(D)} \; 13</math> elements (Pigeonhole principle).
  
 
<i>Solution submitted by green_lotus</i>
 
<i>Solution submitted by green_lotus</i>

Revision as of 10:24, 9 March 2021

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Problem

What is the smallest integer $n$ for which any subset of $\{ 1, 2, 3, \; \dots \; , 20 \}$ of size $n$ must contain two numbers that differ by 8?

$\textbf{(A)} \; 2 \quad \textbf{(B)} \; 8 \quad \textbf{(C)} \; 12 \quad \textbf{(D)} \; 13 \quad \textbf{(E)} \; 15$

Solution

There are twelve pairs $\{ 1, 9 \}$, $\{ 2, 10 \}$, $\{ 3, 11 \}$, $\{ 4, 11 \}$, $\; \dots \; \{ 12, 20 \}$ in $\{ 1, 2, 3, \; \dots \; , 20 \}$ that differ by 8. If we take $n = 12$, it could be that we selected one element from each pair for the subset: the condition may not be fulfilled. the In order to select at least one pair, it is necessary to select $\textbf{(D)} \; 13$ elements (Pigeonhole principle).

Solution submitted by green_lotus

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