The last digit of `9!+3^(9966)` is :

Options

A. 1
B. 3
C. 7
D. 9

Answer

To find the last digit of \(9! + 3^{9966}\), we will break down the problem into two parts: calculating the last digit of \(9!\) and the last digit of \(3^{9966}\).

### Step 1: Calculate the last digit of \(9!\)

The factorial \(9!\) is calculated as follows:
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]

Now, we can observe that \(9!\) contains the factors \(5\) and \(2\). The product of \(5\) and \(2\) gives \(10\), which means that \(9!\) will end with a \(0\). Thus, the last digit of \(9!\) is:
\[ \text{Last digit of } 9! = 0 \]

### Step 2: Calculate the last digit of \(3^{9966}\)

To find the last digit of \(3^{9966}\), we can observe the pattern in the last digits of powers of \(3\):
- \(3^1 = 3\) (last digit is \(3\))
- \(3^2 = 9\) (last digit is \(9\))
- \(3^3 = 27\) (last digit is \(7\))
- \(3^4 = 81\) (last digit is \(1\))
- \(3^5 = 243\) (last digit is \(3\))

The last digits repeat every 4 terms: \(3, 9, 7, 1\).

To determine which last digit corresponds to \(3^{9966}\), we need to find \(9966 \mod 4\):
\[ 9966 \div 4 = 2491 \quad \text{(remainder } 2\text{)} \]
Thus, \(9966 \mod 4 = 2\).

From our pattern, when the exponent is \(2\), the last digit of \(3^{9966}\) is:
\[ \text{Last digit of } 3^{9966} = 9 \]

### Step 3: Combine the results

Now we can add the last digits we found:
\[ \text{Last digit of } 9! + 3^{9966} = 0 + 9 = 9 \]

Thus, the last digit of \(9! + 3^{9966}\) is:
\[ \boxed{9} \]