Find the total number of prime factors in
`2^(17) xx 6^(31) xx 7^(5) xx 10^(11) xx 11^(10) xx (323)^(23)`.

Options

A. `162`
B. `161`
C. `346`
D. `97`

Answer

To find the total number of prime factors in the expression \(2^{17} \times 6^{31} \times 7^{5} \times 10^{11} \times 11^{10} \times (323)^{23}\), we will first break down each term into its prime factors and then sum the powers of these prime factors.

### Step 1: Break down each term into prime factors

1. **For \(2^{17}\)**:
- This is already in prime factor form: \(2^{17}\).

2. **For \(6^{31}\)**:
- \(6 = 2 \times 3\), so:
\[ 6^{31} = (2 \times 3)^{31} = 2^{31} \times 3^{31} \]

3. **For \(7^{5}\)**:
- This is already in prime factor form: \(7^{5}\).

4. **For \(10^{11}\)**:
- \(10 = 2 \times 5\), so:
\[ 10^{11} = (2 \times 5)^{11} = 2^{11} \times 5^{11} \]

5. **For \(11^{10}\)**:
- This is already in prime factor form: \(11^{10}\).

6. **For \(323^{23}\)**:
- First, we need to factor \(323\):
- \(323 = 17 \times 19\) (both are prime numbers), so:
\[ 323^{23} = (17 \times 19)^{23} = 17^{23} \times 19^{23} \]

### Step 2: Combine all the prime factors

Now we combine all the prime factors from each term:

- From \(2^{17}\): \(2^{17}\)
- From \(6^{31}\): \(2^{31} \times 3^{31}\)
- From \(7^{5}\): \(7^{5}\)
- From \(10^{11}\): \(2^{11} \times 5^{11}\)
- From \(11^{10}\): \(11^{10}\)
- From \(323^{23}\): \(17^{23} \times 19^{23}\)

### Step 3: Sum the powers of each prime factor

Now we will sum the powers of each prime factor:

- **For \(2\)**:
\[ 17 + 31 + 11 = 59 \]

- **For \(3\)**:
\[ 31 \]

- **For \(5\)**:
\[ 11 \]

- **For \(7\)**:
\[ 5 \]

- **For \(11\)**:
\[ 10 \]

- **For \(17\)**:
\[ 23 \]

- **For \(19\)**:
\[ 23 \]

### Step 4: Calculate the total number of prime factors

Now we add all the powers together:
\[ 59 + 31 + 11 + 5 + 10 + 23 + 23 = 162 \]

### Final Answer

The total number of prime factors in the expression is **162**.

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