Difference between revisions of "2006 AMC 10A Problems/Problem 20"
Dairyqueenxd (talk | contribs) (→Solution) |
|||
| Line 10: | Line 10: | ||
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>. | 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 == | == See also == | ||
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. 
