2006 AMC 10A Problems/Problem 20

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2006 AMC 10A Problems/Problem 20

Contents

Problem

Solution

Video Solution

See also