Find the last digit of `2^(416)xx4^(430)`.

Answer

To find the last digit of the expression \(2^{416} \times 4^{430}\), we can break it down into manageable steps.

### Step 1: Simplify \(4^{430}\)
We know that \(4\) can be expressed as \(2^2\). Therefore, we can rewrite \(4^{430}\) as:
\[ 4^{430} = (2^2)^{430} = 2^{860} \]
So, the expression \(2^{416} \times 4^{430}\) becomes:
\[ 2^{416} \times 2^{860} = 2^{416 + 860} = 2^{1276} \]

### Step 2: Find the last digit of \(2^{1276}\)
To find the last digit of \(2^{1276}\), we need to observe the pattern in the last digits of the powers of \(2\):
- \(2^1 = 2\) (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)
- \(2^6 = 64\) (last digit is 4)
- \(2^7 = 128\) (last digit is 8)
- \(2^8 = 256\) (last digit is 6)

From this, we can see that the last digits repeat every 4 terms: \(2, 4, 8, 6\).

### Step 3: Determine the position in the cycle
To find out which last digit corresponds to \(2^{1276}\), we need to find \(1276 \mod 4\):
\[ 1276 \div 4 = 319 \quad \text{(exactly, with a remainder of 0)} \]
Thus, \(1276 \mod 4 = 0\).

### Step 4: Find the last digit from the cycle
Since \(1276 \mod 4 = 0\), we refer to the last digit of \(2^4\) in our cycle, which is \(6\).

### Final Answer
Therefore, the last digit of \(2^{416} \times 4^{430}\) is:
\[ \boxed{6} \]