If `p=1!+(2xx2!)+(3xx3!)+….+(10xx10!)`, where `n!=1xx2xx3xx…n` for integer `nge1`, then `p+2` when divisible by 11!, leaves a remainder

Options

A. 10
B. 0
C. 7
D. 1

Answer

To solve the problem, we need to calculate the value of \( p \) given by the expression:

\[ p = 1! + 2 \times 2! + 3 \times 3! + \ldots + 10 \times 10! \]

### Step 1: Rewrite the terms in the sum
We can express \( n \times n! \) in terms of factorials:
\[ n \times n! = n! + (n-1) \times (n-1)! \]
This means we can rewrite each term in the sum as follows:

- \( 1! = 1! \)
- \( 2 \times 2! = 2! + 1! \)
- \( 3 \times 3! = 3! + 2! \)
- \( 4 \times 4! = 4! + 3! \)
- ...
- \( 10 \times 10! = 10! + 9! \)

### Step 2: Expand the sum
Now, we can expand \( p \):
\[ p = 1! + (2! + 1!) + (3! + 2!) + (4! + 3!) + (5! + 4!) + (6! + 5!) + (7! + 6!) + (8! + 7!) + (9! + 8!) + (10! + 9!) \]

### Step 3: Combine like terms
Combining all terms, we can see that:
- Each factorial from \( 1! \) to \( 9! \) appears twice except for \( 10! \), which appears once.
Thus, we can write:
\[ p = 2(1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!) + 10! \]

### Step 4: Calculate the sum of factorials
Now, we need to compute \( 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! \):
\[ 1! = 1 \\ 2! = 2 \\ 3! = 6 \\ 4! = 24 \\ 5! = 120 \\ 6! = 720 \\ 7! = 5040 \\ 8! = 40320 \\ 9! = 362880 \]
Adding these up:
\[ 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 403791 \]

### Step 5: Substitute back into \( p \)
Now substituting back into the expression for \( p \):
\[ p = 2 \times 403791 + 10! = 807582 + 3628800 = 4436382 \]

### Step 6: Calculate \( p + 2 \)
Now we calculate \( p + 2 \):
\[ p + 2 = 4436382 + 2 = 4436384 \]

### Step 7: Find the remainder when dividing by \( 11! \)
Now we need to find \( 11! \):
\[ 11! = 39916800 \]

Now, we calculate the remainder when \( 4436384 \) is divided by \( 39916800 \):
Since \( 4436384 < 39916800 \), the remainder is simply \( 4436384 \).

### Final Answer
Thus, the final answer is:
\[ \text{The remainder when } p + 2 \text{ is divided by } 11! \text{ is } 4436384. \]