The sum of the divisors of `2^3 .3^4 .5^2` is:

Options

A. 15625
B. 1234
C. 56265
D. 56789

Answer

To find the sum of the divisors of the number \( N = 2^3 \cdot 3^4 \cdot 5^2 \), we can use the formula for the sum of the divisors of a number expressed in its prime factorization form. The formula is given by:

\[ \sigma(N) = (p_1^{k_1+1} - 1) / (p_1 - 1) \cdot (p_2^{k_2+1} - 1) / (p_2 - 1) \cdot (p_3^{k_3+1} - 1) / (p_3 - 1) \]

where \( p_1, p_2, p_3 \) are the prime factors and \( k_1, k_2, k_3 \) are their respective powers.

### Step-by-Step Solution:

1. **Identify the Prime Factors and Their Powers**:
- The prime factors of \( N \) are \( 2, 3, \) and \( 5 \).
- Their respective powers are \( 3, 4, \) and \( 2 \).

2. **Apply the Formula for Each Prime Factor**:
- For \( 2^3 \):
\[ \sigma(2^3) = \frac{2^{3+1} - 1}{2 - 1} = \frac{2^4 - 1}{1} = \frac{16 - 1}{1} = 15 \]

- For \( 3^4 \):
\[ \sigma(3^4) = \frac{3^{4+1} - 1}{3 - 1} = \frac{3^5 - 1}{2} = \frac{243 - 1}{2} = \frac{242}{2} = 121 \]

- For \( 5^2 \):
\[ \sigma(5^2) = \frac{5^{2+1} - 1}{5 - 1} = \frac{5^3 - 1}{4} = \frac{125 - 1}{4} = \frac{124}{4} = 31 \]

3. **Multiply the Results**:
- Now, we will multiply the results from each prime factor:
\[ \sigma(N) = \sigma(2^3) \cdot \sigma(3^4) \cdot \sigma(5^2) = 15 \cdot 121 \cdot 31 \]

4. **Calculate the Product**:
- First, calculate \( 15 \cdot 121 \):
\[ 15 \cdot 121 = 1815 \]
- Next, multiply \( 1815 \cdot 31 \):
\[ 1815 \cdot 31 = 56265 \]

5. **Final Result**:
- The sum of the divisors of \( 2^3 \cdot 3^4 \cdot 5^2 \) is \( 56265 \).

### Answer:
The sum of the divisors of \( 2^3 \cdot 3^4 \cdot 5^2 \) is \( 56265 \).

---