# 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 $1 \Longrightarrow \boxed{\mathrm{E}}$.

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