If `(1!)^(2) + (2!)^(2) + (3!)^(2) + "…….." + (99!)^(2)` is divided by `100`, the remainder is

Options

A. 27
B. 28
C. 17
D. 14

Answer

To solve the problem of finding the remainder when the sum \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (99!)^2 \) is divided by 100, we will follow these steps:

### Step 1: Calculate the factorials and their squares
We will compute the squares of the factorials for \( n = 1 \) to \( n = 9 \) first, since for \( n \geq 10 \), \( n! \) will contain at least two factors of 2 and one factor of 5, making \( (n!)^2 \) divisible by 100.

- \( (1!)^2 = 1^2 = 1 \)
- \( (2!)^2 = 2^2 = 4 \)
- \( (3!)^2 = 6^2 = 36 \)
- \( (4!)^2 = 24^2 = 576 \)
- \( (5!)^2 = 120^2 = 14400 \)
- \( (6!)^2 = 720^2 = 518400 \)
- \( (7!)^2 = 5040^2 = 25401600 \)
- \( (8!)^2 = 40320^2 = 1625702400 \)
- \( (9!)^2 = 362880^2 = 131681894400 \)

### Step 2: Calculate the sum of these squares
Now, we will sum the squares calculated above:

\[ S = (1!)^2 + (2!)^2 + (3!)^2 + (4!)^2 + (5!)^2 + (6!)^2 + (7!)^2 + (8!)^2 + (9!)^2 \]

Calculating the sum:

\[ S = 1 + 4 + 36 + 576 + 14400 + 518400 + 25401600 + 1625702400 + 131681894400 \]

Calculating this step-by-step:

- \( 1 + 4 = 5 \)
- \( 5 + 36 = 41 \)
- \( 41 + 576 = 617 \)
- \( 617 + 14400 = 15017 \)
- \( 15017 + 518400 = 533417 \)
- \( 533417 + 25401600 = 25935017 \)
- \( 25935017 + 1625702400 = 1651637417 \)
- \( 1651637417 + 131681894400 = 133333531817 \)

### Step 3: Find the remainder when \( S \) is divided by 100
Now, we need to find the remainder of \( S \) when divided by 100. We can do this by taking \( S \mod 100 \).

Calculating \( 133333531817 \mod 100 \):

The last two digits of \( 133333531817 \) are 17. Therefore,

\[ 133333531817 \mod 100 = 17 \]

### Conclusion
The remainder when \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (99!)^2 \) is divided by 100 is **17**.