2019 AIME II Problems/Problem 3

Problem

Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: $\begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*}$

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

2019 AIME II (Problems • Answer Key • Resources)
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2019 AIME II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

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