`19^(5)+21^(5)` is divisible by

Options

A. only 10
B. only 20
C. both 10 and 20
D. Neither 10 nor 20

Answer

To determine if \( 19^5 + 21^5 \) is divisible by certain numbers, we can use the identity that states if \( n \) is an odd number, then \( a^n + b^n \) is divisible by \( a + b \).

### Step-by-Step Solution:

1. **Identify the values of \( a \) and \( b \)**:
- Here, \( a = 19 \) and \( b = 21 \).

2. **Calculate \( a + b \)**:
- \( a + b = 19 + 21 = 40 \).

3. **Check if \( n \) is odd**:
- Since \( n = 5 \) (which is odd), we can apply the identity.

4. **Apply the identity**:
- According to the identity, \( 19^5 + 21^5 \) is divisible by \( 40 \).

5. **Find the factors of \( 40 \)**:
- The factors of \( 40 \) are \( 1, 2, 4, 5, 8, 10, 20, 40 \).

6. **Conclude the divisibility**:
- Therefore, \( 19^5 + 21^5 \) is divisible by \( 40 \) and all of its factors, which include \( 10 \) and \( 20 \).

### Final Answer:
\( 19^5 + 21^5 \) is divisible by \( 40 \), \( 10 \), and \( 20 \).