Let `y` be an element of the set `A={1,2,3,4,5,6,10,15,30}` and `x_(1)`, `x_(2)`, `x_(3)` be integers such that `x_(1)x_(2)x_(3)=y`, then the number of positive integral solutions of `x_(1)x_(2)x_(3)=y` is

Options

A. `81`
B. `64`
C. `72`
D. `90`

Explanation

To solve the problem of finding the number of positive integral solutions for the equation \( x_1 x_2 x_3 = y \) where \( y \) is an element of the set \( A = \{1, 2, 3, 4, 5, 6, 10, 15, 30\} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the number of positive integral solutions for \( x_1 x_2 x_3 = y \). This means we are looking for combinations of integers \( x_1, x_2, \) and \( x_3 \) that multiply to give \( y \).

2. **Factorization of \( y \)**: For each \( y \) in the set \( A \), we will factor \( y \) into its prime factors. This is important because the number of ways to distribute the prime factors among \( x_1, x_2, \) and \( x_3 \) will give us the number of solutions.

3. **Prime Factorization of Each Element**:
- \( 1 = 1 \) (no prime factors)
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 10 = 2^1 \times 5^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 30 = 2^1 \times 3^1 \times 5^1 \)

4. **Using the Formula for Counting Solutions**: The number of positive integral solutions of the equation \( x_1 x_2 x_3 = y \) can be found using the formula:
\[ \text{Number of solutions} = \frac{(e_1 + k - 1)!}{e_1!(k - 1)!} \times \frac{(e_2 + k - 1)!}{e_2!(k - 1)!} \times \ldots \]
where \( e_i \) are the exponents in the prime factorization of \( y \) and \( k \) is the number of variables (in this case, \( k = 3 \)).

5. **Calculating for Each \( y \)**:
- For \( y = 1 \): No solutions (0).
- For \( y = 2 \): \( 2^1 \) → \( \frac{(1 + 3 - 1)!}{1!(3 - 1)!} = \frac{3!}{1! \cdot 2!} = 3 \).
- For \( y = 3 \): \( 3^1 \) → \( 3 \).
- For \( y = 4 \): \( 2^2 \) → \( \frac{(2 + 3 - 1)!}{2!(3 - 1)!} = \frac{4!}{2! \cdot 2!} = 6 \).
- For \( y = 5 \): \( 5^1 \) → \( 3 \).
- For \( y = 6 \): \( 2^1 \times 3^1 \) → \( \frac{(1 + 3 - 1)!}{1!(3 - 1)!} \cdot \frac{(1 + 3 - 1)!}{1!(3 - 1)!} = 3 \cdot 3 = 9 \).
- For \( y = 10 \): \( 2^1 \times 5^1 \) → \( 9 \).
- For \( y = 15 \): \( 3^1 \times 5^1 \) → \( 9 \).
- For \( y = 30 \): \( 2^1 \times 3^1 \times 5^1 \) → \( 27 \).

6. **Summarizing the Results**: The number of positive integral solutions for each \( y \) is:
- \( y = 1 \): 0
- \( y = 2 \): 3
- \( y = 3 \): 3
- \( y = 4 \): 6
- \( y = 5 \): 3
- \( y = 6 \): 9
- \( y = 10 \): 9
- \( y = 15 \): 9
- \( y = 30 \): 27

### Final Answer:
The number of positive integral solutions for \( x_1 x_2 x_3 = y \) varies depending on the value of \( y \). For \( y = 30 \), the number of solutions is \( 27 \).