Difference between revisions of "2006 AMC 10A Problems/Problem 20"
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== Problem == | == Problem == | ||
− | Six distinct | + | Six distinct positive integers are randomly chosen between <math>1</math> and <math>2006</math>, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of <math>5</math>? |
− | <math>\ | + | <math>\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</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> (their [[remainder]]s after division by 5 must be the same). | For two numbers to have a difference that is a multiple of 5, the numbers must be congruent <math>\bmod{5}</math> (their [[remainder]]s after division by 5 must be the same). |
Revision as of 09:12, 19 December 2021
Problem
Six distinct positive integers are randomly chosen between and , inclusive. What is the probability that some pair of these integers has a difference that is a multiple of ?
Solution
For two numbers to have a difference that is a multiple of 5, the numbers must be congruent (their remainders after division by 5 must be the same).
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 .
See also
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 |
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