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 calculate the last digit of each term separately and then add them together.

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

The factorial \(9!\) is calculated as follows:
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
Notice that within this product, we have the factors \(5\) and \(2\). Multiplying \(5\) and \(2\) gives \(10\), which means that \(9!\) contains \(10\) as a factor. Therefore, the last digit of \(9!\) is \(0\).

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

To find the last digit of \(3^{9966}\), we observe the pattern in the last digits of the 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 \(1\))

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

Next, we need to determine which term in the cycle corresponds to \(3^{9966}\). We do this by calculating \(9966 \mod 4\):
\[ 9966 \div 4 = 2491 \quad \text{remainder} \quad 2 \]
Thus, \(9966 \mod 4 = 2\). This means that \(3^{9966}\) corresponds to the second term in the cycle, which is \(9\).

### Step 3: Combine the results

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

### Final Answer

The last digit of \(9! + 3^{9966}\) is \(9\).

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