Find the remainder when `923^(888) + 235^(222)` is divided by 4.

Answer

To find the remainder when \( 923^{888} + 235^{222} \) is divided by 4, we can follow these steps:

### Step 1: Find the remainder of \( 923 \) when divided by \( 4 \)
To do this, we calculate:
\[ 923 \div 4 = 230 \quad \text{(remainder 3)} \]
So, \( 923 \equiv 3 \mod 4 \).

### Step 2: Find the remainder of \( 235 \) when divided by \( 4 \)
Next, we calculate:
\[ 235 \div 4 = 58 \quad \text{(remainder 3)} \]
Thus, \( 235 \equiv 3 \mod 4 \).

### Step 3: Rewrite the expression using the remainders
Now we can rewrite the original expression:
\[ 923^{888} + 235^{222} \equiv 3^{888} + 3^{222} \mod 4 \]

### Step 4: Determine the power cycle of \( 3 \mod 4 \)
The powers of \( 3 \) modulo \( 4 \) are:
- \( 3^1 \equiv 3 \mod 4 \)
- \( 3^2 \equiv 1 \mod 4 \)
- \( 3^3 \equiv 3 \mod 4 \)
- \( 3^4 \equiv 1 \mod 4 \)

This shows that \( 3^n \mod 4 \) cycles every 2 terms:
- If \( n \) is odd, \( 3^n \equiv 3 \mod 4 \)
- If \( n \) is even, \( 3^n \equiv 1 \mod 4 \)

### Step 5: Determine the parity of the exponents
- \( 888 \) is even, so \( 3^{888} \equiv 1 \mod 4 \)
- \( 222 \) is even, so \( 3^{222} \equiv 1 \mod 4 \)

### Step 6: Combine the results
Now we can combine the results:
\[ 3^{888} + 3^{222} \equiv 1 + 1 \equiv 2 \mod 4 \]

### Step 7: Conclusion
Thus, the remainder when \( 923^{888} + 235^{222} \) is divided by \( 4 \) is:
\[ \boxed{2} \]