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 find the remainder when \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (99!)^2 \) is divided by \( 100 \), we can break the problem down into manageable steps.

### Step 1: Calculate the squares of the factorials for the first few terms
We start by calculating \( (n!)^2 \) for \( n = 1, 2, 3, 4, 5 \).

- \( (1!)^2 = (1)^2 = 1 \)
- \( (2!)^2 = (2)^2 = 4 \)
- \( (3!)^2 = (6)^2 = 36 \)
- \( (4!)^2 = (24)^2 = 576 \)
- \( (5!)^2 = (120)^2 = 14400 \)

### Step 2: Find the contributions of \( (n!)^2 \) for \( n \geq 5 \)
For \( n \geq 5 \), \( n! \) contains at least two factors of \( 2 \) and two factors of \( 5 \), which means \( n! \) is divisible by \( 100 \). Therefore, \( (n!)^2 \) for \( n \geq 5 \) will also be divisible by \( 100 \) and will contribute \( 0 \) to the sum when taken modulo \( 100 \).

### Step 3: Sum the contributions from \( n = 1 \) to \( n = 4 \)
Now, we sum the contributions from \( n = 1 \) to \( n = 4 \):

\[ (1!)^2 + (2!)^2 + (3!)^2 + (4!)^2 = 1 + 4 + 36 + 576 \]

Calculating this gives:

\[ 1 + 4 = 5 \]
\[ 5 + 36 = 41 \]
\[ 41 + 576 = 617 \]

### Step 4: Calculate the remainder when divided by \( 100 \)
Now we need to find the remainder of \( 617 \) when divided by \( 100 \):

\[ 617 \div 100 = 6 \quad \text{(quotient)} \]
\[ 617 - (6 \times 100) = 617 - 600 = 17 \]

Thus, the remainder when \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (99!)^2 \) is divided by \( 100 \) is \( 17 \).

### Final Answer
The remainder is \( \boxed{17} \).