When x is divided by 4 leaves remainder 3 then `(2022+x)^2022` is divisided by 8 , remainder is

Answer

To solve the problem, we need to find the remainder when \((2022 + x)^{2022}\) is divided by 8, given that \(x\) leaves a remainder of 3 when divided by 4.

### Step-by-step Solution:

1. **Express \(x\) in terms of \(k\)**:
Since \(x\) leaves a remainder of 3 when divided by 4, we can express \(x\) as:
\[ x = 4k + 3 \]
for some integer \(k\).

2. **Substitute \(x\) into the expression**:
Now, substitute \(x\) into the expression \(2022 + x\):
\[ 2022 + x = 2022 + (4k + 3) = 2025 + 4k \]

3. **Simplify modulo 8**:
Next, we need to find \(2025 + 4k\) modulo 8. First, we calculate \(2025 \mod 8\):
\[ 2025 \div 8 = 253 \quad \text{(integer part)} \]
\[ 2025 - (253 \times 8) = 2025 - 2024 = 1 \]
Thus,
\[ 2025 \equiv 1 \mod 8 \]
Therefore,
\[ 2025 + 4k \equiv 1 + 4k \mod 8 \]

4. **Consider \(4k \mod 8\)**:
The term \(4k\) can take values \(0\) or \(4\) depending on whether \(k\) is even or odd:
- If \(k\) is even, \(4k \equiv 0 \mod 8\).
- If \(k\) is odd, \(4k \equiv 4 \mod 8\).

5. **Combine results**:
Therefore:
- If \(k\) is even:
\[ 1 + 4k \equiv 1 + 0 \equiv 1 \mod 8 \]
- If \(k\) is odd:
\[ 1 + 4k \equiv 1 + 4 \equiv 5 \mod 8 \]

6. **Raise to the power of 2022**:
Now we need to find \((1 + 4k)^{2022} \mod 8\):
- If \(k\) is even:
\[ (1)^{2022} \equiv 1 \mod 8 \]
- If \(k\) is odd:
\[ (5)^{2022} \mod 8 \]
To find \(5^{2022} \mod 8\), we can observe the pattern:
\[ 5^1 \equiv 5 \mod 8 \]
\[ 5^2 \equiv 25 \equiv 1 \mod 8 \]
Since \(5^2 \equiv 1 \mod 8\), all even powers of 5 will also be congruent to 1:
\[ 5^{2022} \equiv 1 \mod 8 \]

7. **Conclusion**:
In both cases (whether \(k\) is even or odd), we find that:
\[ (2022 + x)^{2022} \equiv 1 \mod 8 \]
Therefore, the remainder when \((2022 + x)^{2022}\) is divided by 8 is:
\[ \boxed{1} \]