AIME 2020 I · 第 9 题

Solution 1

First, prime factorize $20^9$ as $2^{18} \cdot 5^9$. Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$, $a_2$ as $2^{b_2} \cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \cdot 5^{c_3}$.

In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\le b_2\le b_3$, and $c_1\le c_2\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor.

We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\le b_1. Therefore, if we pick$3$integers from$0$to$20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers$b_1$,$b_2+1$, and$b_3+2$. 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$\dbinom{21}{3}$.

The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is

$$\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.$$ Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.

-molocyxu

Solution 2

Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we must have $b_1 \leq b_2 \leq b_3$. If they are all distinct, the number of cases is simply ${19 \choose 3}$. If they are all equal, there are only $19$ cases for the general value. If we have a pair equal, then we have $2 \cdot {19\choose 2}$. We need to multiply by $2$ because if we have two values $b_i < b_j$, we can have either $(b_i, b_i, b_j)$ or $(b_i, b_j, b_j)$.

$${19 \choose 3} + 2 \cdot {19 \choose 2} + 19 = 1330$$ Likewise for $c$, we get

$${10 \choose 3} + 2 \cdot {10 \choose 2} + 10 = 220$$ The final probability is simply $\frac{1330 \cdot 220}{190^3}$. Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.

Solution 3

Similar to before, we calculate that there are $190^3$ 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 $\dbinom{12}{3}$ ways to choose $d$, $f$ and $g$, and $\dbinom{21}{3}$ ways to choose $a$, $b$, and $c$. Therefore, the probability of choosing $3$ factors which satisfy the conditions is

$$\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.$$ This simplifies to $\frac{77}{1805}$, therefore $m =$ $\boxed{077}$.

Solution 4 (Official MAA)

Note that the prime factorization of $20^9$ is $2^{18}\cdot5^{9}.$ The problem reduces to selecting independently the powers of $2$ and the powers of $5$ in the numbers $a_1$, $a_2$, and $a_3$. This is equivalent to selecting $3$ exponents for the powers of $2$ and $3$ exponents for the powers of $5$ and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer $k$, the probability that three integers $a$, $b$, and $c$ are chosen such that $0\le a \le b\le c \le k$ is the probability that $a$, $b+1$, and $c+2$ are chosen such that $0 \le a < b+1 < c+2 \le k+2$. There are $\binom {k+3}3$ ways to choose $a$, $b+1$, and $c+2$, so the probability that the integers are chosen in order is

$$\frac{\binom{k+3}3}{(k+1)^3}.$$ Thus the probability that three chosen divisors of $20^9$ satisfy the divisibility requirement is

$$\frac{\binom{12}3}{10^3}\cdot\frac{\binom{21}3}{19^3}=\frac{12\cdot11\cdot10}{6\cdot10\cdot10\cdot10}\cdot\frac{21\cdot20\cdot19}{6\cdot19\cdot19\cdot19}=\frac{77}{1805}.$$ The requested numerator is $77.$