If n is the number of positive integral solutions of `x_1 x_2 x_3 x_4 = 210`. Then

Options

A. n must be divisible by 3 distinct primes
B. n must be a perfect square
C. n msut be a perfect 4th power
D. n must be a perfect 8th power

Answer

To find the number of positive integral solutions for the equation \( x_1 x_2 x_3 x_4 = 210 \), we will follow these steps:

### Step 1: Factor the number 210
First, we need to find the prime factorization of 210.

\[ 210 = 2 \times 3 \times 5 \times 7 \]

### Step 2: Write the equation in terms of prime factors
We can express \( x_1, x_2, x_3, \) and \( x_4 \) in terms of the prime factors:

\[ x_1 = 2^{a_1} \times 3^{b_1} \times 5^{c_1} \times 7^{d_1} \]
\[ x_2 = 2^{a_2} \times 3^{b_2} \times 5^{c_2} \times 7^{d_2} \]
\[ x_3 = 2^{a_3} \times 3^{b_3} \times 5^{c_3} \times 7^{d_3} \]
\[ x_4 = 2^{a_4} \times 3^{b_4} \times 5^{c_4} \times 7^{d_4} \]

### Step 3: Set up the equations for the exponents
From the equation \( x_1 x_2 x_3 x_4 = 210 \), we can equate the exponents of the prime factors:

\[ a_1 + a_2 + a_3 + a_4 = 1 \quad \text{(for the prime 2)} \]
\[ b_1 + b_2 + b_3 + b_4 = 1 \quad \text{(for the prime 3)} \]
\[ c_1 + c_2 + c_3 + c_4 = 1 \quad \text{(for the prime 5)} \]
\[ d_1 + d_2 + d_3 + d_4 = 1 \quad \text{(for the prime 7)} \]

### Step 4: Use the stars and bars method
The number of non-negative integral solutions to each of these equations can be found using the stars and bars theorem. The number of solutions to the equation \( x_1 + x_2 + x_3 + x_4 = k \) is given by:

\[ \binom{n+k-1}{k} \]

where \( n \) is the number of variables (in our case, 4) and \( k \) is the sum (in our case, 1).

For each equation, we have:

\[ \text{Number of solutions for } a_1 + a_2 + a_3 + a_4 = 1 = \binom{4-1}{1} = \binom{3}{1} = 3 \]

This is true for each of the four equations (for \( b, c, \) and \( d \) as well).

### Step 5: Calculate the total number of solutions
Since the equations are independent, we multiply the number of solutions for each prime factor:

\[ n = 3 \times 3 \times 3 \times 3 = 3^4 = 81 \]

### Conclusion
Thus, the number of positive integral solutions \( n \) for the equation \( x_1 x_2 x_3 x_4 = 210 \) is:

\[ \boxed{81} \]