Difference between revisions of "2020 AIME I Problems/Problem 9"
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Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math> | Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math> | ||
− | == Solution == | + | == Solution 1 == |
<asy> | <asy> | ||
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==Solution 3== | ==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/ | + | Similar to before, we calculate that there are <math>190^3</math> ways to choose <math>3</math> factors with replacement. Then, we figure out the number of triplets <math>{a,b,c}</math> and <math>{d,f,g}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> represent powers of <math>2</math> and <math>d</math>, <math>f</math>, and <math>g</math> represent powers of <math>5</math>, such that the triplets are in non-descending order. The maximum power of <math>2</math> is <math>18</math>, and the maximum power of <math>5</math> is <math>9</math>. Using the Hockey Stick identity, we figure out that there are <math>\dbinom{12}{3}</math> ways to choose <math>d</math>, <math>f</math> and <math>g</math>, and <math>\dbinom{21}{3}</math> ways to choose <math>a</math>, <math>b</math>, and <math>c</math>. Therefore, the probability of choosing <math>3</math> factors which satisfy the conditions is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> This simplifies to <math>\frac{77}{1805}</math>, therefore <math>m =</math> <math>\boxed{077}</math>. |
+ | |||
+ | ==Solution 4 (Official MAA)== | ||
+ | Note that the prime factorization of <math>20^9</math> is <math>2^{18}\cdot5^{9}.</math> The problem reduces to selecting independently the powers of <math>2</math> and the powers of <math>5</math> in the numbers <math>a_1</math>, <math>a_2</math>, and <math>a_3</math>. This is equivalent to selecting <math>3</math> exponents for the powers of <math>2</math> and <math>3</math> exponents for the powers of <math>5</math> and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer <math>k</math>, the probability that three integers <math>a</math>, <math>b</math>, and <math>c</math> are chosen such that <math>0\le a \le b\le c \le k</math> is the probability that <math>a</math>, <math>b+1</math>, and <math>c+2</math> are chosen such that <math>0 \le a < b+1 < c+2 \le k+2</math>. There are <math>\binom {k+3}3</math> ways to choose <math>a</math>, <math>b+1</math>, and <math>c+2</math>, so the probability that the integers are chosen in order is | ||
+ | <cmath>\frac{\binom{k+3}3}{(k+1)^3}.</cmath>Thus the probability that three chosen divisors of <math>20^9</math> satisfy the divisibility requirement is | ||
+ | <cmath>\frac{\binom{12}3}{10^3}\cdot\frac{\binom{21}3}{19^3}=\frac{12\cdot11\cdot10}{6\cdot10\cdot10\cdot10}\cdot\frac{21\cdot20\cdot19}{6\cdot19\cdot19\cdot19}=\frac{77}{1805}.</cmath>The requested numerator is <math>77.</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=8|num-a=10}} | {{AIME box|year=2020|n=I|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category: Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:59, 25 February 2021
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 1
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
factors with replacement. Then, we figure out the number of triplets
and
, where
,
, and
represent powers of
and
,
, and
represent powers of
, such that the triplets are in non-descending order. The maximum power of
is
, and the maximum power of
is
. Using the Hockey Stick identity, we figure out that there are
ways to choose
,
and
, and
ways to choose
,
, and
. Therefore, the probability of choosing
factors which satisfy the conditions is
This simplifies to
, therefore
.
Solution 4 (Official MAA)
Note that the prime factorization of is
The problem reduces to selecting independently the powers of
and the powers of
in the numbers
,
, and
. This is equivalent to selecting
exponents for the powers of
and
exponents for the powers of
and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer
, the probability that three integers
,
, and
are chosen such that
is the probability that
,
, and
are chosen such that
. There are
ways to choose
,
, and
, so the probability that the integers are chosen in order is
Thus the probability that three chosen divisors of
satisfy the divisibility requirement is
The requested numerator is
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 |
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