The unit digit of the following expression
`(1 !)^(99) + (2!)^(98) + (3!)^(97) + (4!)^(96) + …..(99 !)^1` is :

Options

A. 1
B. 3
C. 7
D. 6

Answer

To find the unit digit of the expression

\[ (1!)^{99} + (2!)^{98} + (3!)^{97} + (4!)^{96} + \ldots + (99!)^1, \]

we will analyze the unit digits of each factorial raised to its respective power.

### Step 1: Calculate the unit digits of factorials

1. **Calculate \(1!\)**:
\[ 1! = 1 \quad \text{(unit digit is 1)} \]

2. **Calculate \(2!\)**:
\[ 2! = 2 \quad \text{(unit digit is 2)} \]

3. **Calculate \(3!\)**:
\[ 3! = 6 \quad \text{(unit digit is 6)} \]

4. **Calculate \(4!\)**:
\[ 4! = 24 \quad \text{(unit digit is 4)} \]

5. **Calculate \(5!\)**:
\[ 5! = 120 \quad \text{(unit digit is 0)} \]

6. **Calculate \(n!\) for \(n \geq 5\)**:
For \(n \geq 5\), \(n!\) will always have a unit digit of 0 because it includes the factors 2 and 5, which multiply to give 10.

### Step 2: Analyze the powers

Now we consider the powers of these factorials:

- **For \( (1!)^{99} \)**:
\[ (1!)^{99} = 1^{99} = 1 \quad \text{(unit digit is 1)} \]

- **For \( (2!)^{98} \)**:
\[ (2!)^{98} = 2^{98} \quad \text{(unit digit is 2)} \]

- **For \( (3!)^{97} \)**:
\[ (3!)^{97} = 6^{97} \quad \text{(unit digit is 6)} \]

- **For \( (4!)^{96} \)**:
\[ (4!)^{96} = 4^{96} \quad \text{(unit digit is 6, since \(4^1 = 4\), \(4^2 = 16\), \(4^3 = 64\), \(4^4 = 256\) and repeats every 2)} \]

- **For \( (5!)^{95} \)** and all higher factorials:
\[ (5!)^{95} = 0^{95} = 0 \quad \text{(unit digit is 0)} \]
Similarly, \( (n!)^k \) for \( n \geq 5 \) will also have a unit digit of 0.

### Step 3: Sum the unit digits

Now we sum the unit digits from \( (1!)^{99} \) to \( (4!)^{96} \):

\[ \text{Unit digit of } (1!)^{99} + (2!)^{98} + (3!)^{97} + (4!)^{96} + (5!)^{95} + \ldots + (99!)^1 \]

Calculating the unit digits:
- From \( (1!)^{99} \): 1
- From \( (2!)^{98} \): 2
- From \( (3!)^{97} \): 6
- From \( (4!)^{96} \): 6
- From \( (5!)^{95} \) and higher: 0

So, we have:
\[ 1 + 2 + 6 + 6 + 0 + 0 + \ldots + 0 = 15 \]

### Step 4: Find the unit digit of the sum

The unit digit of 15 is:
\[ \text{Unit digit is } 5. \]

### Conclusion

The unit digit of the entire expression is **5**.

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