The remainder when `(1!)^(2) + (2!)^(2) + (3!)^(2) + ….. + (100!)^(2)` is divided by `144` is `ab` then `(a)/(b) =`

Explanation

To solve the problem, we need to find the remainder when the sum \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (100!)^2 \) is divided by 144. Let's break this down step by step.

### Step 1: Calculate the factorials and their squares for the first few terms
- \( (1!)^2 = (1)^2 = 1 \)
- \( (2!)^2 = (2)^2 = 4 \)
- \( (3!)^2 = (6)^2 = 36 \)
- \( (4!)^2 = (24)^2 = 576 \)

### Step 2: Determine the contribution of \( (n!)^2 \) for \( n \geq 4 \)
For \( n \geq 4 \), \( n! \) becomes larger, and we can see that \( n! \) will be divisible by 144. This is because:
- \( 4! = 24 \) and \( 24^2 = 576 \) is divisible by 144.
- For \( n \geq 5 \), \( n! \) includes at least two factors of 2 and at least one factor of 3, making \( (n!)^2 \) divisible by \( 144 \).

### Step 3: Calculate the total contribution from \( n = 1, 2, 3 \)
Now, we only need to consider the contributions from \( n = 1, 2, 3 \):
\[ (1!)^2 + (2!)^2 + (3!)^2 = 1 + 4 + 36 = 41 \]

### Step 4: Find the remainder when divided by 144
Since \( (n!)^2 \) for \( n \geq 4 \) contributes 0 to the remainder when divided by 144, the total sum modulo 144 is:
\[ 41 \mod 144 = 41 \]

### Step 5: Identify \( a \) and \( b \)
We are given that the remainder can be expressed as \( ab \), where \( a = 4 \) and \( b = 1 \) (since \( 41 = 4 \times 10 + 1 \)). Thus, we have:
- \( a = 4 \)
- \( b = 1 \)

### Step 6: Calculate \( \frac{a}{b} \)
Now, we calculate \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{4}{1} = 4 \]

### Final Answer
Therefore, the final answer is:
\[ \frac{a}{b} = 4 \]