If `N=1!+2!+3!-4!+….+47!-48!+49!`, then what is the unit digit of `N^(N)` ?

Options

A. 0
B. 9
C. 7
D. 1

Answer

To solve the problem step by step, we need to compute the value of \( N = 1! + 2! + 3! - 4! + \ldots + 47! - 48! + 49! \) and then find the unit digit of \( N^N \).

### Step 1: Calculate the factorial values for the first few numbers
1. \( 1! = 1 \)
2. \( 2! = 2 \)
3. \( 3! = 6 \)
4. \( 4! = 24 \)
5. \( 5! = 120 \) (and notice that the unit digit is 0)
6. \( 6! = 720 \) (unit digit is 0)
7. For \( n \geq 5 \), \( n! \) will always have a unit digit of 0.

### Step 2: Sum the relevant factorials
Since \( 5! \) and all higher factorials contribute a unit digit of 0, we only need to consider \( 1!, 2!, 3!, \) and \( 4! \).

Now, we can compute \( N \):
\[ N = 1! + 2! + 3! - 4! = 1 + 2 + 6 - 24 \]

### Step 3: Calculate the sum
Now we perform the arithmetic:
\[ N = 1 + 2 + 6 - 24 = 9 - 24 = -15 \]

### Step 4: Find the unit digit of \( N \)
The unit digit of \( N = -15 \) is 5 (since we only care about the unit digit).

### Step 5: Calculate \( N^N \)
Now we need to find the unit digit of \( N^N \) where \( N \) has a unit digit of 5.

### Step 6: Determine the unit digit of \( 5^N \)
The unit digit of any power of 5 is always 5. Therefore, the unit digit of \( N^N \) is:
\[ \text{Unit digit of } N^N = 5 \]

### Final Answer
Thus, the unit digit of \( N^N \) is **5**.

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