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

m (2006 AMC 10A Problem 20 moved to 2006 AMC 10A Problems/Problem 20) |
(Added solution) |
||

Line 4: | Line 4: | ||

<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> | ||

+ | |||

+ | <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>. | ||

+ | |||

+ | 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 21:43, 17 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?

## Solution

For two numbers to have a difference that is a multiple of 5, the numbers must be congruent

are the possible values of numbers in .

Since there are only 5 possible values in and we are picking numbers, by the Pigeonhole Principle, two of the numbers must be congruent .

Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is