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

Answer

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:
\[ \boxed{60} \]