PRMO 2017 question 21 - factors of 108

21. Find the number of ordered triples (a, b, c) of positive integers such that abc = 108.
Prerequisite:

Stars and bars theorem.

Solution:

To find the number of ordered triples $(a, b, c)$ of positive integers such that $abc = 108$, we can follow these steps:

Step 1: Prime Factorization of 108

First, we need to factor 108 into its prime factors.

$$108 = 2^2 \times 3^3$$

Step 2: Assigning Prime Factors to $a$, $b$, and $c$

We can express $a$, $b$, and $c$ in terms of their prime factors:

$$a = 2^r \times 3^q,\quad b = 2^t \times 3^u,\quad c = 2^z \times 3^v$$

where $r, t, z$ are the powers of 2 and $q, u, v$ are the powers of 3.

Step 3: Setting Up the Equations

From the product $abc = 108$, we can equate the powers of the prime factors:

$$r + t + z = 2\quad \text{(for the power of 2)}$$ $$q + u + v = 3\quad \text{(for the power of 3)}$$

Step 4: Finding Non-Negative Integer Solutions

We need to find the number of non-negative integer solutions to the equations.

For $r + t + z = 2$:

Using the "stars and bars" theorem, the number of solutions is given by:

$$\text{Number of solutions} = \binom{n + k - 1}{k - 1}$$

where $n$ is the total we want (2 in this case) and $k$ is the number of variables (3 here: $r, t, z$).

$$\text{Number of solutions} = \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6$$

For $q + u + v = 3$:

Similarly, we apply the same formula:

$$\text{Number of solutions} = \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

Step 5: Total Number of Ordered Triples

Since the distributions of the powers of 2 and 3 are independent, we multiply the number of solutions:

$$\text{Total ordered triples} = 6 \times 10 = 60$$

Final Answer

Thus, the number of ordered triples $(a, b, c)$ such that $abc = 108$ is:

$$\boxed{{\displaystyle 60}}$$