If `S=1!+4!+7!+10!+ . . .+400!,` then
Q. The last two digits in the number S is divisible by

Options

A. 13
B. 12
C. 11
D. 10

Answer

To solve the problem \( S = 1! + 4! + 7! + 10! + \ldots + 400! \) and find the last two digits of \( S \) that are divisible by a certain number, we can follow these steps:

### Step 1: Identify the terms in the series
The series consists of factorials of numbers that follow the pattern \( 1, 4, 7, 10, \ldots, 400 \). This is an arithmetic sequence where the first term \( a = 1 \) and the common difference \( d = 3 \).

### Step 2: Determine the last two digits of each factorial
- **1! = 1** (last two digits: 01)
- **4! = 24** (last two digits: 24)
- **7! = 5040** (last two digits: 40)
- **10! = 3628800** (last two digits: 00)

From \( 10! \) onwards, every factorial will have at least two trailing zeros because \( 10! \) contains the factors \( 2 \) and \( 5 \) multiple times. Thus, the last two digits of \( n! \) for \( n \geq 10 \) will be \( 00 \).

### Step 3: Sum the last two digits of the relevant factorials
Now we only need to sum the last two digits of \( 1! \), \( 4! \), and \( 7! \):
\[ S = 1! + 4! + 7! = 1 + 24 + 40 \]
Calculating this gives:
\[ S = 1 + 24 + 40 = 65 \]

### Step 4: Check divisibility of the last two digits
Now we need to find out if \( 65 \) is divisible by any number. The prime factorization of \( 65 \) is \( 5 \times 13 \). Thus, \( 65 \) is divisible by \( 1, 5, 13, \) and \( 65 \).

### Conclusion
The last two digits of \( S \) are \( 65 \), and the number \( 65 \) is divisible by \( 13 \).

### Final Answer
The last two digits in the number \( S \) is divisible by \( 13 \).

---