Difference between revisions of "2006 AMC 10A Problems/Problem 20"
m (added category and link to previous and next problem) |
|||
| (6 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
== 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? | + | 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> | + | For two numbers to have a difference that is a multiple of <math>5</math>, the numbers must be congruent <math>\bmod{5}</math> (their remainders after division by <math>5</math> must be the same). |
| − | |||
| − | <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>. | + | <math>0, 1, 2, 3, 4 </math> are the possible values of numbers in <math>\bmod{5}</math>. Since there are only <math>5</math> 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> | + | Therefore the probability that some pair of the <math>6</math> integers has a difference that is a multiple of <math>5</math> is <math>\boxed{\textbf{(E) }1}</math>. |
| − | == | + | ==Video Solution== |
| − | + | https://youtu.be/jfkW_KwI9Wo | |
| − | + | ~savannahsolver | |
| − | + | == See also == | |
| + | {{AMC10 box|year=2006|ab=A|num-b=19|num-a=21}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:21, 17 March 2023
Contents
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 , the numbers must be congruent 
(their remainders after division by must be the same).
are the possible values of numbers in . Since there are only 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 integers has a difference that is a multiple of is 
.
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
