Digit at the units place of sum of
`(1!)^(2)+(2!)^(2)+(3!)^(2).......+(2008!)^(2)` is

Options

A. 5
B. 7
C. 9
D. 6

Answer

To find the digit at the units place of the sum \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (2008!)^2 \), we can follow these steps:

### Step 1: Calculate the units digit of the factorial squares for small values of \( n \).

1. **For \( n = 1 \)**:
\[ (1!)^2 = (1)^2 = 1 \]
Units digit = 1

2. **For \( n = 2 \)**:
\[ (2!)^2 = (2)^2 = 4 \]
Units digit = 4

3. **For \( n = 3 \)**:
\[ (3!)^2 = (6)^2 = 36 \]
Units digit = 6

4. **For \( n = 4 \)**:
\[ (4!)^2 = (24)^2 = 576 \]
Units digit = 6

5. **For \( n = 5 \)**:
\[ (5!)^2 = (120)^2 = 14400 \]
Units digit = 0

6. **For \( n = 6 \)**:
\[ (6!)^2 = (720)^2 = 518400 \]
Units digit = 0

### Step 2: Identify the pattern for \( n \geq 5 \).

From \( n = 5 \) onwards, the factorial \( n! \) will always include the factors 2 and 5, which means \( n! \) will end with a 0. Therefore, for \( n \geq 5 \):
\[ (n!)^2 \text{ will also end with a 0.} \]
Thus, the units digit of \( (n!)^2 \) for \( n \geq 5 \) is 0.

### Step 3: Sum the units digits for \( n = 1 \) to \( n = 4 \).

Now, we can sum the units digits we calculated:
\[ \text{Units digit of } (1!)^2 = 1 \]
\[ \text{Units digit of } (2!)^2 = 4 \]
\[ \text{Units digit of } (3!)^2 = 6 \]
\[ \text{Units digit of } (4!)^2 = 6 \]

Adding these:
\[ 1 + 4 + 6 + 6 = 17 \]

### Step 4: Find the units digit of the total sum.

The units digit of 17 is:
\[ \text{Units digit} = 7 \]

### Conclusion

Thus, the digit at the units place of the sum \( (1!)^2 + (2!)^2 + (3!)^2 + \ldots + (2008!)^2 \) is **7**.

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