The remainder when `(2345)^(676)+(1234)^(567)` is divided by 4 is _____________

Answer

To find the remainder when \( (2345)^{676} + (1234)^{567} \) is divided by 4, we can simplify the problem using modular arithmetic.

### Step-by-Step Solution:

1. **Find \( 2345 \mod 4 \)**:
\[ 2345 \div 4 = 586 \quad \text{(quotient)} \]
\[ 2345 - (586 \times 4) = 2345 - 2344 = 1 \]
Thus, \( 2345 \equiv 1 \mod 4 \).

2. **Find \( (2345)^{676} \mod 4 \)**:
Since \( 2345 \equiv 1 \mod 4 \):
\[ (2345)^{676} \equiv 1^{676} \equiv 1 \mod 4 \]

3. **Find \( 1234 \mod 4 \)**:
\[ 1234 \div 4 = 308 \quad \text{(quotient)} \]
\[ 1234 - (308 \times 4) = 1234 - 1232 = 2 \]
Thus, \( 1234 \equiv 2 \mod 4 \).

4. **Find \( (1234)^{567} \mod 4 \)**:
Since \( 1234 \equiv 2 \mod 4 \):
\[ (1234)^{567} \equiv 2^{567} \mod 4 \]
We know that:
- \( 2^1 \equiv 2 \mod 4 \)
- \( 2^2 \equiv 0 \mod 4 \)
- For any \( n \geq 2 \), \( 2^n \equiv 0 \mod 4 \).

Since \( 567 \geq 2 \):
\[ 2^{567} \equiv 0 \mod 4 \]

5. **Combine results**:
Now we can combine the results:
\[ (2345)^{676} + (1234)^{567} \equiv 1 + 0 \equiv 1 \mod 4 \]

Thus, the remainder when \( (2345)^{676} + (1234)^{567} \) is divided by 4 is **1**.