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Difference between revisions of "2006 AMC 10A Problems/Problem 20"

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== Problem ==
 
== Problem ==
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Six distinct [[positive]] [[integer]]s are randomly chosen between 1 and 2006, inclusive. What is the [[probability]] that some pair of these integers has a difference that is a multiple of 5?  
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Six distinct positive integers are randomly chosen between <math>1</math> and <math>2006</math>, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of <math>5</math>?  
 
   
 
   
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<math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{3}{5}\qquad\mathrm{(C) \ } \frac{2}{3}\qquad\mathrm{(D) \ } \frac{4}{5}\qquad\mathrm{(E) \ } 1\qquad</math>
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<math>\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{3}{5}\qquad\textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{(E) } 1\qquad</math>
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== Solution ==
 
== Solution ==
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For two numbers to have a difference that is a multiple of 5, the numbers must be congruent <math>\bmod{5}</math> (their [[remainder]]s after division by 5 must be the same).
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For two numbers to have a difference that is a multiple of <math>5</math>, the numbers must be congruent <math>\bmod{5}</math> (their remainders after division by <math>5</math> must be the same).
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<math>0, 1, 2, 3, 4 </math> are the possible values of numbers in <math>\bmod{5}</math>. Since there are only <math>5</math> possible values in <math>\bmod{5}</math> and we are picking <math>6</math> numbers, by the [[Pigeonhole Principle]], two of the numbers must be congruent <math>\bmod{5}</math>.
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Therefore the probability that some pair of the <math>6</math> integers has a difference that is a multiple of <math>5</math> is <math>\boxed{\textbf{(E) }1}</math>.
  
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<math>0, 1, 2, 3, 4 </math> are the possible values of numbers in <math>\bmod{5}</math>. Since there are only 5 possible values in <math>\bmod{5}</math> and we are picking <math>6</math> numbers, by the [[Pigeonhole Principle]], two of the numbers must be congruent <math>\bmod{5}</math>.
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==Video Solution==
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https://youtu.be/jfkW_KwI9Wo
  
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Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is <math>1 \Longrightarrow \mathrm{E}</math>.
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~savannahsolver
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2006|ab=A|num-b=17|num-a=19}}
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{{AMC10 box|year=2006|ab=A|num-b=19|num-a=21}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 07:21, 17 March 2023

Problem

Six distinct positive integers are randomly chosen between $1$ and $2006$, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$?

$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{3}{5}\qquad\textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{(E) } 1\qquad$

Solution

For two numbers to have a difference that is a multiple of $5$, the numbers must be congruent $\bmod{5}$ (their remainders after division by $5$ must be the same).

$0, 1, 2, 3, 4$ are the possible values of numbers in $\bmod{5}$. Since there are only $5$ possible values in $\bmod{5}$ and we are picking $6$ numbers, by the Pigeonhole Principle, two of the numbers must be congruent $\bmod{5}$.

Therefore the probability that some pair of the $6$ integers has a difference that is a multiple of $5$ is $\boxed{\textbf{(E) }1}$.

Video Solution

https://youtu.be/jfkW_KwI9Wo

~savannahsolver

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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