Find the unit digit in the expression :`(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ......................... + (100!)^(100!)`

Options

A. (a) 7
B. (b) 4
C. (c) 6
D. (d) 8

Answer

To find the unit digit of the expression \( (1!)^{(1!)} + (2!)^{(2!)} + (3!)^{(3!)} + \ldots + (100!)^{(100!)} \), we can break it down step by step.

### Step 1: Calculate the factorials and their powers for the first few terms

1. **Calculate \( 1! \)**:
\[ 1! = 1 \]
So, \( (1!)^{(1!)} = 1^1 = 1 \).

2. **Calculate \( 2! \)**:
\[ 2! = 2 \]
So, \( (2!)^{(2!)} = 2^2 = 4 \).

3. **Calculate \( 3! \)**:
\[ 3! = 6 \]
So, \( (3!)^{(3!)} = 6^6 \). To find the unit digit of \( 6^6 \), we note that the unit digit of any power of 6 is always 6.

4. **Calculate \( 4! \)**:
\[ 4! = 24 \]
So, \( (4!)^{(4!)} = 24^{24} \). The unit digit of \( 24 \) is \( 4 \). The unit digits of powers of \( 4 \) cycle through \( 4, 6 \). Since \( 24 \mod 2 = 0 \), the unit digit of \( 24^{24} \) is \( 6 \).

### Step 2: Analyze higher factorials

For \( n \geq 5 \):
- \( n! \) will always have a unit digit of \( 0 \) because \( n! \) includes the factors \( 2 \) and \( 5 \), which multiply to give \( 10 \). Therefore, \( (n!)^{(n!)} \) for \( n \geq 5 \) will also have a unit digit of \( 0 \).

### Step 3: Sum the unit digits

Now we can sum the unit digits from \( n = 1 \) to \( n = 4 \):
- From \( n = 1 \): unit digit is \( 1 \)
- From \( n = 2 \): unit digit is \( 4 \)
- From \( n = 3 \): unit digit is \( 6 \)
- From \( n = 4 \): unit digit is \( 6 \)
- From \( n = 5 \) to \( n = 100 \): unit digit is \( 0 \)

Thus, the total unit digit is:
\[ 1 + 4 + 6 + 6 = 17 \]

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

The unit digit of \( 17 \) is \( 7 \).

### Final Answer

The unit digit of the expression \( (1!)^{(1!)} + (2!)^{(2!)} + (3!)^{(3!)} + \ldots + (100!)^{(100!)} \) is **7**.