Consider the number `N=2016`.Find the Sum of the even divivisors of the number N.

Options

A. a) 6552
B. 6448
C. 6048
D. 5733

Explanation

To find the sum of the even divisors of the number \( N = 2016 \), we can follow these steps:

### Step 1: Prime Factorization of 2016
First, we need to factor the number 2016 into its prime factors.

\[ 2016 = 2^5 \times 3^2 \times 7^1 \]

### Step 2: Identify the Even Divisors
Even divisors of a number must include at least one factor of 2. Therefore, we can express the even divisors of 2016 in the form:

\[ \text{Even Divisors} = 2^k \times 3^m \times 7^n \]

where \( k \) can take values from 1 to 5 (since we need at least one factor of 2), \( m \) can take values from 0 to 2, and \( n \) can take values from 0 to 1.

### Step 3: Calculate the Sum of Even Divisors
The sum of the even divisors can be calculated using the formula for the sum of divisors based on prime factorization:

\[ \sigma(N) = (p_1^{k_1+1} - 1)/(p_1 - 1) \times (p_2^{k_2+1} - 1)/(p_2 - 1) \times \ldots \]

For our case, we need to calculate the sum of the even divisors:

1. **Sum of powers of 2** (from \( 2^1 \) to \( 2^5 \)):
\[ 2 + 2^2 + 2^3 + 2^4 + 2^5 = 2 + 4 + 8 + 16 + 32 = 62 \]

2. **Sum of powers of 3** (from \( 3^0 \) to \( 3^2 \)):
\[ 3^0 + 3^1 + 3^2 = 1 + 3 + 9 = 13 \]

3. **Sum of powers of 7** (from \( 7^0 \) to \( 7^1 \)):
\[ 7^0 + 7^1 = 1 + 7 = 8 \]

### Step 4: Combine the Results
Now, we can combine these sums to find the total sum of the even divisors:

\[ \text{Sum of Even Divisors} = (62) \times (13) \times (8) \]

Calculating this gives:

\[ 62 \times 13 = 806 \]
\[ 806 \times 8 = 6448 \]

### Final Answer
Thus, the sum of the even divisors of the number \( N = 2016 \) is:

\[ \boxed{6448} \]