When `(43^101 + 23^101)` is divided by 66, then the remainder will be?

Options

A. 1
B. 2
C. 3
D. 0

Answer

To find the remainder when \( 43^{101} + 23^{101} \) is divided by 66, we can use properties of modular arithmetic.

### Step-by-step Solution:

1. **Identify the Expression**: We need to evaluate \( 43^{101} + 23^{101} \mod 66 \).

2. **Use the Chinese Remainder Theorem**: Since 66 can be factored into \( 2 \times 3 \times 11 \), we will find \( 43^{101} + 23^{101} \) modulo 2, 3, and 11 separately.

3. **Calculate \( \mod 2 \)**:
- \( 43 \equiv 1 \mod 2 \)
- \( 23 \equiv 1 \mod 2 \)
- Therefore, \( 43^{101} + 23^{101} \equiv 1^{101} + 1^{101} \equiv 1 + 1 \equiv 0 \mod 2 \).

4. **Calculate \( \mod 3 \)**:
- \( 43 \equiv 1 \mod 3 \)
- \( 23 \equiv 2 \mod 3 \)
- Therefore, \( 43^{101} + 23^{101} \equiv 1^{101} + 2^{101} \equiv 1 + 2 \equiv 0 \mod 3 \).

5. **Calculate \( \mod 11 \)**:
- \( 43 \equiv 10 \mod 11 \)
- \( 23 \equiv 1 \mod 11 \)
- Therefore, \( 43^{101} + 23^{101} \equiv 10^{101} + 1^{101} \).
- By Fermat's Little Theorem, \( 10^{10} \equiv 1 \mod 11 \), so \( 10^{101} = (10^{10})^{10} \cdot 10^1 \equiv 1^{10} \cdot 10 \equiv 10 \mod 11 \).
- Thus, \( 10^{101} + 1^{101} \equiv 10 + 1 \equiv 11 \equiv 0 \mod 11 \).

6. **Combine Results Using CRT**:
- We have:
- \( 43^{101} + 23^{101} \equiv 0 \mod 2 \)
- \( 43^{101} + 23^{101} \equiv 0 \mod 3 \)
- \( 43^{101} + 23^{101} \equiv 0 \mod 11 \)
- Since the results are all 0, we conclude that \( 43^{101} + 23^{101} \equiv 0 \mod 66 \).

7. **Final Result**: The remainder when \( 43^{101} + 23^{101} \) is divided by 66 is **0**.