Difference between revisions of "2020 AIME I Problems/Problem 9"
Line 61: | Line 61: | ||
The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>. | The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Similar to before, we calculate that there are <math>190^3</math> ways to choose 3 factors with replacement. Then, we figure out the number of triplets {a,b,c} and {d,f,g}, where a, b, and c represent powers of 2 and d, f, and g represent powers of 5, such that the triplets are in non-descending order. The maximum power of 2 is 18, and the maximum power of 5 is 9. Using the Hockey Stick identity, we figure out that there are <math>\dbinom{12}{3}</math> ways to choose d, f and g, and <math>\dbinom{21}{3}</math> ways to choose a, b, and c. Therefore, the probability of choosing 3 factors which satisfy the conditions is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> This simplifies to 77/1805, therefore m = <math>\boxed{077}</math>. | ||
==See Also== | ==See Also== |
Revision as of 11:04, 29 March 2020
Contents
[hide]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 between the numbers
,
, and
. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. 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
Simplification gives
, and therefore the answer is
.
-molocyxu
Solution 2
Same as before, say the factors have powers of and
.
can either be all distinct, all equal, or two of the three are equal. As well, we must have
. If they are all distinct, the number of cases is simply
. If they are all equal, there are only
cases for the general value. If we have a pair equal, then we have
. We need to multiply by
because if we have two values
, we can have either
or
.
Likewise for , we get
The final probability is simply . Simplification gives
, and therefore the answer is
.
Solution 3
Similar to before, we calculate that there are ways to choose 3 factors with replacement. Then, we figure out the number of triplets {a,b,c} and {d,f,g}, where a, b, and c represent powers of 2 and d, f, and g represent powers of 5, such that the triplets are in non-descending order. The maximum power of 2 is 18, and the maximum power of 5 is 9. Using the Hockey Stick identity, we figure out that there are
ways to choose d, f and g, and
ways to choose a, b, and c. Therefore, the probability of choosing 3 factors which satisfy the conditions is
This simplifies to 77/1805, therefore m =
.
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.