The total number of divisors of the number `N=2^(5).3^(4).5^(10).7^(6)` that are of the form `4K+2, AAK in N` is equal to

Options

A. 385
B. 384
C. 96
D. 77

Answer

To find the total number of divisors of the number \( N = 2^5 \cdot 3^4 \cdot 5^{10} \cdot 7^6 \) that are of the form \( 4k + 2 \), we can follow these steps:

### Step 1: Understanding the Form \( 4k + 2 \)
A number of the form \( 4k + 2 \) can be expressed as \( 2 \times \text{(odd number)} \). This means that any divisor of \( N \) that is of the form \( 4k + 2 \) must include exactly one factor of \( 2 \) and the remaining factors must come from the odd part of \( N \).

### Step 2: Identify the Odd Part of \( N \)
The odd part of \( N \) is obtained by ignoring the factor of \( 2^5 \). Thus, the odd part is:
\[ N_{odd} = 3^4 \cdot 5^{10} \cdot 7^6 \]

### Step 3: Count the Odd Divisors
To find the number of odd divisors of \( N_{odd} \), we use the formula for the number of divisors. If \( N = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k} \), then the number of divisors \( d(N) \) is given by:
\[ d(N) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]

For \( N_{odd} = 3^4 \cdot 5^{10} \cdot 7^6 \):
- The exponent of \( 3 \) is \( 4 \), so \( e_1 + 1 = 4 + 1 = 5 \)
- The exponent of \( 5 \) is \( 10 \), so \( e_2 + 1 = 10 + 1 = 11 \)
- The exponent of \( 7 \) is \( 6 \), so \( e_3 + 1 = 6 + 1 = 7 \)

Thus, the total number of odd divisors is:
\[ d(N_{odd}) = 5 \cdot 11 \cdot 7 \]

### Step 4: Calculate the Total Number of Odd Divisors
Now we can compute:
\[ 5 \cdot 11 = 55 \]
\[ 55 \cdot 7 = 385 \]

### Step 5: Form the Divisors of the Form \( 4k + 2 \)
Since we need divisors of the form \( 4k + 2 \), we multiply the number of odd divisors by \( 1 \) (since we are including exactly one factor of \( 2 \)):
\[ \text{Total divisors of the form } 4k + 2 = 385 \]

### Final Answer
The total number of divisors of \( N \) that are of the form \( 4k + 2 \) is \( 385 \).