The sum of all the proper divisors of 9900 is

Options

A. 23951
B. 23952
C. 23953
D. none of these

Answer

To find the sum of all proper divisors of 9900, we will follow these steps:

### Step 1: Factorization of 9900
First, we need to factor 9900 into its prime factors.

9900 can be factored as:
\[ 9900 = 2^2 \times 3^2 \times 5^2 \times 11^1 \]

### Step 2: Use the Sum of Divisors Formula
The formula for the sum of divisors \( \sigma(n) \) for a number \( n = p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k} \) is given by:
\[ \sigma(n) = (p_1^{0} + p_1^{1} + \ldots + p_1^{a_1})(p_2^{0} + p_2^{1} + \ldots + p_2^{a_2}) \ldots (p_k^{0} + p_k^{1} + \ldots + p_k^{a_k}) \]

### Step 3: Calculate Each Term
Now we will calculate each term for our prime factorization:

1. For \( 2^2 \):
\[ 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7 \]

2. For \( 3^2 \):
\[ 3^0 + 3^1 + 3^2 = 1 + 3 + 9 = 13 \]

3. For \( 5^2 \):
\[ 5^0 + 5^1 + 5^2 = 1 + 5 + 25 = 31 \]

4. For \( 11^1 \):
\[ 11^0 + 11^1 = 1 + 11 = 12 \]

### Step 4: Multiply the Results
Now we will multiply these results together to find the sum of all divisors:
\[ \sigma(9900) = 7 \times 13 \times 31 \times 12 \]

Calculating step-by-step:
1. \( 7 \times 13 = 91 \)
2. \( 91 \times 31 = 2821 \)
3. \( 2821 \times 12 = 33852 \)

So, the sum of all divisors of 9900 is:
\[ \sigma(9900) = 33852 \]

### Step 5: Calculate the Sum of Proper Divisors
The sum of proper divisors is given by:
\[ \text{Sum of Proper Divisors} = \sigma(9900) - 9900 - 1 \]
Substituting the values:
\[ \text{Sum of Proper Divisors} = 33852 - 9900 - 1 = 23951 \]

### Final Answer
The sum of all proper divisors of 9900 is:
\[ \boxed{23951} \]