Find the last digit of `222^(888) + 388^(222)`.

Answer

To find the last digit of \( 222^{888} + 388^{222} \), we can follow these steps:

### Step 1: Identify the last digits
First, we need to find the last digits of \( 222^{888} \) and \( 388^{222} \). The last digit of a number is determined by its units place.

- The last digit of \( 222 \) is \( 2 \).
- The last digit of \( 388 \) is \( 8 \).

Thus, we need to calculate \( 2^{888} \) and \( 8^{222} \).

### Step 2: Calculate the last digit of \( 2^{888} \)
The last digits of powers of \( 2 \) cycle every 4:
- \( 2^1 = 2 \) (last digit is 2)
- \( 2^2 = 4 \) (last digit is 4)
- \( 2^3 = 8 \) (last digit is 8)
- \( 2^4 = 16 \) (last digit is 6)
- \( 2^5 = 32 \) (last digit is 2) and so on.

The cycle is \( 2, 4, 8, 6 \).

To find \( 2^{888} \), we need to find \( 888 \mod 4 \):
\[ 888 \div 4 = 222 \quad \text{(remainder 0)} \]
Since the remainder is \( 0 \), we take the last digit from the 4th position in the cycle, which is \( 6 \).

### Step 3: Calculate the last digit of \( 8^{222} \)
The last digits of powers of \( 8 \) also cycle every 4:
- \( 8^1 = 8 \) (last digit is 8)
- \( 8^2 = 64 \) (last digit is 4)
- \( 8^3 = 512 \) (last digit is 2)
- \( 8^4 = 4096 \) (last digit is 6)
- \( 8^5 = 32768 \) (last digit is 8) and so on.

The cycle is \( 8, 4, 2, 6 \).

To find \( 8^{222} \), we need to find \( 222 \mod 4 \):
\[ 222 \div 4 = 55 \quad \text{(remainder 2)} \]
Since the remainder is \( 2 \), we take the last digit from the 2nd position in the cycle, which is \( 4 \).

### Step 4: Add the last digits
Now we add the last digits obtained from the previous steps:
\[ 6 + 4 = 10 \]
The last digit of \( 10 \) is \( 0 \).

### Final Answer
Thus, the last digit of \( 222^{888} + 388^{222} \) is \( 0 \).

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