When `(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ……+(100!)^(100!)` is divided by 5, the remainder obtained is :

Options

A. 2
B. 0
C. 4
D. none of these

Answer

To solve the problem of finding the remainder when \( (1!)^{(1!)} + (2!)^{(2!)} + (3!)^{(3!)} + \ldots + (100!)^{(100!)} \) is divided by 5, we can break it down step by step.

### Step 1: Calculate the factorials and their powers

1. **Calculate \( (1!)^{(1!)} \)**:
\[ 1! = 1 \quad \Rightarrow \quad (1!)^{(1!)} = 1^1 = 1 \]

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

3. **Calculate \( (3!)^{(3!)} \)**:
\[ 3! = 6 \quad \Rightarrow \quad (3!)^{(3!)} = 6^6 \]

4. **Calculate \( (4!)^{(4!)} \)**:
\[ 4! = 24 \quad \Rightarrow \quad (4!)^{(4!)} = 24^{24} \]

5. **Calculate \( (5!)^{(5!)} \)**:
\[ 5! = 120 \quad \Rightarrow \quad (5!)^{(5!)} = 120^{120} \]

### Step 2: Analyze the powers modulo 5

Next, we need to find the value of each term modulo 5.

1. **For \( (1!)^{(1!)} \mod 5 \)**:
\[ 1 \mod 5 = 1 \]

2. **For \( (2!)^{(2!)} \mod 5 \)**:
\[ 4 \mod 5 = 4 \]

3. **For \( (3!)^{(3!)} \mod 5 \)**:
- Calculate \( 6^6 \mod 5 \):
\[ 6 \mod 5 = 1 \quad \Rightarrow \quad 6^6 \mod 5 = 1^6 = 1 \]

4. **For \( (4!)^{(4!)} \mod 5 \)**:
- Calculate \( 24^{24} \mod 5 \):
\[ 24 \mod 5 = 4 \quad \Rightarrow \quad 4^{24} \mod 5 \]
- Since \( 4 \equiv -1 \mod 5 \):
\[ (-1)^{24} \mod 5 = 1 \]

5. **For \( (5!)^{(5!)} \mod 5 \)**:
- Since \( 120 \equiv 0 \mod 5 \):
\[ 120^{120} \mod 5 = 0 \]

### Step 3: Combine the results

Now we can combine the results of all the terms calculated modulo 5:
\[ (1) + (4) + (1) + (1) + (0) + \ldots + (0) \quad \text{(for terms from } 6! \text{ to } 100!) \]

The terms from \( 6! \) to \( 100! \) will all contribute \( 0 \) because they will all be multiples of 5.

So, we only need to sum the non-zero contributions:
\[ 1 + 4 + 1 + 1 = 7 \]

### Step 4: Find the remainder when divided by 5

Finally, we find the remainder when 7 is divided by 5:
\[ 7 \mod 5 = 2 \]

### Final Answer
The remainder obtained when \( (1!)^{(1!)} + (2!)^{(2!)} + (3!)^{(3!)} + \ldots + (100!)^{(100!)} \) is divided by 5 is **2**.

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