Difference between revisions of "2006 AMC 10A Problems/Problem 20"
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<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>. | ||
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| + | 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 
