Find the number of positive integral solutions of `x y z=21600.`

Explanation

To find the number of positive integral solutions of the equation \(xyz = 21600\), we will follow these steps:

### Step 1: Prime Factorization of 21600
First, we need to factor 21600 into its prime factors.

1. Start dividing by the smallest prime number, which is 2:
- \(21600 \div 2 = 10800\)
- \(10800 \div 2 = 5400\)
- \(5400 \div 2 = 2700\)
- \(2700 \div 2 = 1350\)
- \(1350 \div 2 = 675\) (no longer divisible by 2)

2. Now, switch to the next prime number, which is 3:
- \(675 \div 3 = 225\)
- \(225 \div 3 = 75\)
- \(75 \div 3 = 25\) (no longer divisible by 3)

3. Finally, divide by 5:
- \(25 \div 5 = 5\)
- \(5 \div 5 = 1\)

Thus, the prime factorization of \(21600\) is:
\[ 21600 = 2^5 \times 3^3 \times 5^2 \]

### Step 2: Finding the Number of Positive Integral Solutions
The number of positive integral solutions for the equation \(xyz = n\) can be determined using the formula:
\[ \text{Number of solutions} = (e_1 + 1)(e_2 + 1)(e_3 + 1) \]
where \(e_1, e_2, e_3\) are the powers of the prime factors.

From the prime factorization:
- The power of 2 is \(5\) (i.e., \(e_1 = 5\))
- The power of 3 is \(3\) (i.e., \(e_2 = 3\))
- The power of 5 is \(2\) (i.e., \(e_3 = 2\))

Now we can substitute these values into the formula:
\[ \text{Number of solutions} = (5 + 1)(3 + 1)(2 + 1) \]

Calculating each term:
- \(5 + 1 = 6\)
- \(3 + 1 = 4\)
- \(2 + 1 = 3\)

Now multiply these results:
\[ \text{Number of solutions} = 6 \times 4 \times 3 \]

Calculating the product:
\[ 6 \times 4 = 24 \]
\[ 24 \times 3 = 72 \]

### Final Answer
Thus, the number of positive integral solutions of \(xyz = 21600\) is \(72\).

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