Difference between revisions of "2006 AMC 10A Problems/Problem 20"

 
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== Problem ==
 
== 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?  
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Six distinct [[positive]] [[integer]]s 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?  
 
   
 
   
 
<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 ==
== See Also ==
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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).
*[[2006 AMC 10A Problems]]
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<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 \Longrightarrow \boxed{\mathrm{E}}</math>.
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== See also ==
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{{AMC10 box|year=2006|ab=A|num-b=19|num-a=21}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 16:14, 22 April 2021

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?

$\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{3}{5}\qquad\mathrm{(C) \ } \frac{2}{3}\qquad\mathrm{(D) \ } \frac{4}{5}\qquad\mathrm{(E) \ } 1\qquad$

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

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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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