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Difference between revisions of "2002 AMC 10P Problems/Problem 15"
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== Solution ==
 
== Solution ==
 
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. By the Pigeonhole principle, in order to select at least one pair, it is necessary to select <math>\boxed{\textbf{(D) } 13}</math> elements.
 
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. By the Pigeonhole principle, in order to select at least one pair, it is necessary to select <math>\boxed{\textbf{(D) } 13}</math> elements.
 +
 
~green_lotus
 
~green_lotus
  

Revision as of 07:44, 15 July 2024

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. By the Pigeonhole principle, in order to select at least one pair, it is necessary to select $\boxed{\textbf{(D) } 13}$ elements.

~green_lotus

See Also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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