AIME 2019 II · 第 3 题

Solution 1

As 71 is prime, $c$, $d$, and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$. Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$, which separates easily into two subproblems. The number of positive integer solutions to $ab = 70$ simply equals the number of divisors of 70 (as we can choose a divisor for $a$, which uniquely determines $b$). As $70 = 2^1 \cdot 5^1 \cdot 7^1$, we have $d(70) = (1+1)(1+1)(1+1) = 8$ solutions. Similarly, $72 = 2^3 \cdot 3^2$, so $d(72) = 4 \times 3 = 12$.

Then the answer is simply $8 \times 12 = \boxed{096}$.

-scrabbler94

Solution 2

We know that any two consecutive numbers are coprime. Using this, we can figure out that $c=1$ and $e=1$. $d$ then has to be $71$. Now we have two equations left. $ab=70$ and $fg=72$. To solve these we just need to figure out all of the factors. Doing the prime factorization of $70$ and $72$, we find that they have $8$ and $12$ factors, respectively. The answer is $8 \times 12=\boxed{096}$

~kempwood