The remainder obtained when `1!+2!+3!+……+11!` is divided by 12 is :

Options

A. 9
B. 8
C. 7
D. 6

Answer

To solve the problem of finding the remainder when \(1! + 2! + 3! + \ldots + 11!\) is divided by 12, we can follow these steps:

### Step 1: Calculate the factorials from \(1!\) to \(11!\)

1. \(1! = 1\)
2. \(2! = 2\)
3. \(3! = 6\)
4. \(4! = 24\)
5. \(5! = 120\)
6. \(6! = 720\)
7. \(7! = 5040\)
8. \(8! = 40320\)
9. \(9! = 362880\)
10. \(10! = 3628800\)
11. \(11! = 39916800\)

### Step 2: Identify which factorials contribute to the sum modulo 12

Since \(4! = 24\) and all higher factorials (i.e., \(5!, 6!, \ldots, 11!\)) are multiples of 12, they will contribute a remainder of 0 when divided by 12. Therefore, we only need to consider \(1!, 2!, 3!\), and \(4!\).

### Step 3: Calculate the sum of the relevant factorials

Now, we sum the factorials we identified:
\[ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 \]

### Step 4: Find the remainder when the sum is divided by 12

Now, we need to find the remainder of \(33\) when divided by \(12\):
\[ 33 \div 12 = 2 \quad \text{(quotient)} \]
\[ 33 - (12 \times 2) = 33 - 24 = 9 \]

### Final Answer

The remainder when \(1! + 2! + 3! + \ldots + 11!\) is divided by \(12\) is \(9\).