What is the total number of factors of the number `N=4^(11)xx14^(5)xx11^(2)?`

Answer

To find the total number of factors of the number \( N = 4^{11} \times 14^{5} \times 11^{2} \), we will first express \( N \) in terms of its prime factors and then use the formula for finding the total number of factors.

### Step 1: Break down the components into prime factors

1. **Break down \( 4^{11} \)**:
\[ 4 = 2^2 \implies 4^{11} = (2^2)^{11} = 2^{22} \]

2. **Break down \( 14^{5} \)**:
\[ 14 = 2 \times 7 \implies 14^{5} = (2 \times 7)^{5} = 2^{5} \times 7^{5} \]

3. **Break down \( 11^{2} \)**:
\[ 11^{2} \text{ is already in prime factor form.} \]

### Step 2: Combine all the prime factors

Now, we can combine all the prime factors:
\[ N = 2^{22} \times (2^{5} \times 7^{5}) \times 11^{2} \]

### Step 3: Combine the powers of the same base

1. Combine the powers of \( 2 \):
\[ 2^{22} \times 2^{5} = 2^{22 + 5} = 2^{27} \]

2. So, we can rewrite \( N \) as:
\[ N = 2^{27} \times 7^{5} \times 11^{2} \]

### Step 4: Use the formula to find the total number of factors

The formula to find the total number of factors of a number \( N = p_1^{e_1} \times p_2^{e_2} \times p_3^{e_3} \) is:
\[ \text{Total factors} = (e_1 + 1)(e_2 + 1)(e_3 + 1) \]

For our number \( N = 2^{27} \times 7^{5} \times 11^{2} \):
- \( e_1 = 27 \) (for \( 2 \))
- \( e_2 = 5 \) (for \( 7 \))
- \( e_3 = 2 \) (for \( 11 \))

Now, substituting the values into the formula:
\[ \text{Total factors} = (27 + 1)(5 + 1)(2 + 1) = 28 \times 6 \times 3 \]

### Step 5: Calculate the total number of factors

1. Calculate \( 28 \times 6 \):
\[ 28 \times 6 = 168 \]

2. Now, calculate \( 168 \times 3 \):
\[ 168 \times 3 = 504 \]

### Final Answer

Thus, the total number of factors of the number \( N \) is \( 504 \).

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