Difference between revisions of "2020 AIME I Problems/Problem 9"
(→Problem) |
Kevinmathz (talk | contribs) |
||
| Line 1: | Line 1: | ||
| − | |||
== Problem == | == Problem == | ||
Revision as of 16:47, 12 March 2020
Problem
Let 
be the set of positive integer divisors of 
Three numbers are chosen independently and at random with replacement from the set and labeled 
and 
in the order they are chosen. The probability that both 
divides 
and divides is 
where 
and 
are relatively prime positive integers. Find 
Solution
First, prime factorize 
as 
. Denote as 
, as 
, and as 
.
In order for to divide , and for to divide , 
, and 
. We will consider each case separately. Note that the total amount of possibilities is 
, as there are 
choices for each factor.
We notice that if we add 
to 
and 
to 
, then we can reach the stronger inequality 
. Therefore, if we pick 
integers from 
to 
, they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is 
.
The case for 
,
, and 
proceeds similarly for a result of 
. Therefore, the probability of choosing three such factors is ![\[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\]](http://latex.artofproblemsolving.com/8/1/7/817cd42178e0b531ab9f976e36422ecf8aff72e1.png)
Simplification gives 
, and therefore the answer is 
.
-molocyxu
See Also
| 2020 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
