Difference between revisions of "2006 AMC 10A Problems/Problem 20"
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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>. | <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>. | + | 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>. |
== See also == | == See also == |
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?
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 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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