The number of positive integral solutions of the equation `x_(1)x_(2)x_(3)=60` is:

Options

A. 54
B. 27
C. 81
D. none

Answer

To find the number of positive integral solutions of the equation \( x_1 x_2 x_3 = 60 \), we can follow these steps:

### Step 1: Prime Factorization of 60
First, we need to perform the prime factorization of 60.
\[ 60 = 2^2 \times 3^1 \times 5^1 \]

### Step 2: Expressing Variables in Terms of Prime Factors
We can express \( x_1, x_2, \) and \( x_3 \) in terms of the prime factors:
\[ x_1 = 2^{a_1} \times 3^{b_1} \times 5^{c_1} \]
\[ x_2 = 2^{a_2} \times 3^{b_2} \times 5^{c_2} \]
\[ x_3 = 2^{a_3} \times 3^{b_3} \times 5^{c_3} \]

### Step 3: Setting Up the Equations
From the product \( x_1 x_2 x_3 = 60 \), we can derive the following equations based on the powers of the prime factors:
1. For \( 2 \): \( a_1 + a_2 + a_3 = 2 \)
2. For \( 3 \): \( b_1 + b_2 + b_3 = 1 \)
3. For \( 5 \): \( c_1 + c_2 + c_3 = 1 \)

### Step 4: Finding the Number of Non-negative Integer Solutions
We can use the "stars and bars" theorem to find the number of non-negative integer solutions for each equation.

#### For \( a_1 + a_2 + a_3 = 2 \):
Using the formula for the number of solutions \( \binom{n + r - 1}{r - 1} \):
\[ n = 2, \quad r = 3 \quad \Rightarrow \quad \text{Number of solutions} = \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6 \]

#### For \( b_1 + b_2 + b_3 = 1 \):
\[ n = 1, \quad r = 3 \quad \Rightarrow \quad \text{Number of solutions} = \binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3 \]

#### For \( c_1 + c_2 + c_3 = 1 \):
\[ n = 1, \quad r = 3 \quad \Rightarrow \quad \text{Number of solutions} = \binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3 \]

### Step 5: Total Number of Positive Integral Solutions
Now, we multiply the number of solutions for each equation to find the total number of positive integral solutions:
\[ \text{Total Solutions} = 6 \times 3 \times 3 = 54 \]

### Final Answer
Thus, the number of positive integral solutions of the equation \( x_1 x_2 x_3 = 60 \) is \( \boxed{54} \).

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