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 $\boxed{\textbf{(E) }1}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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