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

Options

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

Answer

To find the unit digit of the expression \( (1!)^{99} + (2!)^{98} + (3!)^{97} + \ldots + (99!)^{1} \), we will calculate the unit digit of each term in the series and then sum them up, focusing only on the unit digits.

### Step-by-Step Solution:

1. **Calculate Unit Digits of Factorials:**
- \( 1! = 1 \) → Unit digit = 1
- \( 2! = 2 \) → Unit digit = 2
- \( 3! = 6 \) → Unit digit = 6
- \( 4! = 24 \) → Unit digit = 4
- \( 5! = 120 \) → Unit digit = 0
- For \( n \geq 5 \), \( n! \) will always have a unit digit of 0 because it contains the factors 2 and 5.

2. **Evaluate Each Term:**
- \( (1!)^{99} = 1^{99} \) → Unit digit = 1
- \( (2!)^{98} = 2^{98} \) → The unit digit of \( 2^{98} \) can be found by observing the pattern of unit digits of powers of 2: 2, 4, 8, 6 (repeats every 4). Since \( 98 \mod 4 = 2 \), the unit digit is 4.
- \( (3!)^{97} = 6^{97} \) → The unit digit of \( 6^n \) is always 6 for \( n \geq 1 \).
- \( (4!)^{96} = 4^{96} \) → The unit digit of \( 4^n \) follows the pattern: 4, 6 (repeats every 2). Since \( 96 \mod 2 = 0 \), the unit digit is 6.
- For \( n \geq 5 \), \( (n!)^{(100-n)} \) will have a unit digit of 0.

3. **Sum the Unit Digits:**
- From the calculations:
- \( (1!)^{99} \) → 1
- \( (2!)^{98} \) → 4
- \( (3!)^{97} \) → 6
- \( (4!)^{96} \) → 6
- \( (5!)^{95} \) to \( (99!)^{1} \) → All contribute 0.
- Therefore, the total unit digit is:
\[ 1 + 4 + 6 + 6 + 0 + 0 + \ldots + 0 = 17 \]
- The unit digit of 17 is **7**.

### Final Answer:
The unit digit of the expression \( (1!)^{99} + (2!)^{98} + \ldots + (99!)^{1} \) is **7**.