Difference between revisions of "2006 AMC 10A Problems/Problem 20"
Dairyqueenxd (talk | contribs) (→Problem) |
|||
(One intermediate revision by one other user not shown) | |||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | For two numbers to have a difference that is a multiple of 5, the numbers must be congruent <math>\bmod{5}</math> (their | + | 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>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>. | + | <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 == | == See also == |
Latest revision as of 07:21, 17 March 2023
Contents
[hide]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.