The sum of the divisors of 9600 is

Options

A. 3048
B. 6120
C. 31620
D. 24384

Answer

To find the sum of the divisors of 9600, we will follow these steps:

### Step 1: Prime Factorization of 9600
First, we need to express 9600 as a product of its prime factors.

1. Start by breaking down 9600:
\[ 9600 = 96 \times 100 \]

2. Factor each component:
- For \( 96 \):
\[ 96 = 2^5 \times 3^1 \]
- For \( 100 \):
\[ 100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2 \]

3. Combine the factors:
\[ 9600 = 2^5 \times 3^1 \times 2^2 \times 5^2 = 2^{5+2} \times 3^1 \times 5^2 = 2^7 \times 3^1 \times 5^2 \]

### Step 2: Use the Formula for the Sum of Divisors
The formula for the sum of the divisors \( \sigma(n) \) of a number \( n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_m^{k_m} \) is given by:
\[ \sigma(n) = (1 + p_1 + p_1^2 + \ldots + p_1^{k_1})(1 + p_2 + p_2^2 + \ldots + p_2^{k_2}) \ldots (1 + p_m + p_m^2 + \ldots + p_m^{k_m}) \]

For \( 9600 = 2^7 \times 3^1 \times 5^2 \), we apply the formula:

1. Calculate the sum for \( 2^7 \):
\[ 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = \frac{2^{8} - 1}{2 - 1} = 255 \]

2. Calculate the sum for \( 3^1 \):
\[ 1 + 3 = 4 \]

3. Calculate the sum for \( 5^2 \):
\[ 1 + 5 + 5^2 = 1 + 5 + 25 = 31 \]

### Step 3: Multiply the Results
Now, we multiply the results from each prime factor:
\[ \sigma(9600) = 255 \times 4 \times 31 \]

1. First, calculate \( 255 \times 4 \):
\[ 255 \times 4 = 1020 \]

2. Then, calculate \( 1020 \times 31 \):
\[ 1020 \times 31 = 31620 \]

### Final Result
Thus, the sum of the divisors of 9600 is:
\[ \boxed{31620} \]