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

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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>
 
<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>
 
== Solution ==
 
== Solution ==
For two numbers to have a difference that is a multiple of 5, the numbers must be congruent <math>\bmod{5}</math>
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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>.
  
 
<math> 0 , 1, 2, 3, 4 </math> are the possible values of numbers in <math>\bmod{5}</math>.
 
<math> 0 , 1, 2, 3, 4 </math> are the possible values of numbers in <math>\bmod{5}</math>.
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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>.  
 
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>.  
  
Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is <math>1 \Rightarrow E </math>
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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 \Rightarrow E </math>.
  
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]

Revision as of 12:15, 18 July 2006

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?

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

Solution

For two numbers to have a difference that is a multiple of 5, the numbers must be congruent $\bmod{5}$.

$0 , 1, 2, 3, 4$ 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 $1 \Rightarrow E$.

See Also

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