Difference between revisions of "2006 AMC 10A Problems/Problem 20"
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== Problem == | == Problem == | ||
| − | Six distinct positive | + | 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 == | ||
| − | For two numbers to have a difference that is a multiple of 5, the numbers must be congruent <math>\bmod{5}</math>. | + | 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). |
| − | <math> 0 , 1, 2, 3, 4 </math> are the possible values of numbers in <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>. |
| − | + | Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is <math>1 \Longrightarrow \mathrm{E}</math>. | |
| − | + | == See also == | |
| − | + | {{AMC10 box|year=2006|ab=A|num-b=17|num-a=19}} | |
| − | == See | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 16:19, 26 February 2007
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 
(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 17 |
Followed by Problem 19 | |
| 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 | ||
