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 factor 9600 into its prime factors.

1. **Divide by 10**:
\( 9600 = 960 \times 10 \)
\( 10 = 2 \times 5 \)
So, we have \( 9600 = 960 \times 2 \times 5 \).

2. **Factor 960**:
\( 960 = 96 \times 10 \)
Again, \( 10 = 2 \times 5 \), so we have \( 960 = 96 \times 2 \times 5 \).

3. **Factor 96**:
\( 96 = 32 \times 3 \)
\( 32 = 2^5 \), so \( 96 = 2^5 \times 3^1 \).

Putting it all together:
- \( 9600 = 2^5 \times 3^1 \times 2^1 \times 5^1 \)
- Combine the powers of 2:
\( 9600 = 2^{6} \times 3^{1} \times 5^{2} \)

Thus, the prime factorization of 9600 is:
\[ 9600 = 2^6 \times 3^1 \times 5^2 \]

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

For our case:
- \( p_1 = 2, k_1 = 6 \)
- \( p_2 = 3, k_2 = 1 \)
- \( p_3 = 5, k_3 = 2 \)

### Step 3: Calculate Each Part
1. **For \( 2^6 \)**:
\[ 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127 \]

2. **For \( 3^1 \)**:
\[ 1 + 3 = 4 \]

3. **For \( 5^2 \)**:
\[ 1 + 5 + 5^2 = 1 + 5 + 25 = 31 \]

### Step 4: Multiply the Results
Now we multiply these results together:
\[ \sigma(9600) = 127 \times 4 \times 31 \]

Calculating this step by step:
1. \( 127 \times 4 = 508 \)
2. \( 508 \times 31 = 15748 \)

Thus, the sum of the divisors of 9600 is:
\[ \boxed{15748} \]