The factorial of a number n is exactly divisible by `(2^(11) xx 11^2)` then the least possible value of n is :

Options

A. 22
B. 25
C. dose not exist
D. none of these

Answer

To solve the problem of finding the least possible value of \( n \) such that \( n! \) is exactly divisible by \( 2^{11} \times 11^2 \), we will follow these steps:

### Step 1: Understand the requirements for divisibility
We need \( n! \) to be divisible by \( 2^{11} \) and \( 11^2 \). This means we need to ensure that \( n! \) contains at least 11 factors of 2 and at least 2 factors of 11.

### Step 2: Finding the minimum \( n \) for \( 11^2 \)
To find the minimum \( n \) such that \( n! \) contains at least 2 factors of 11, we can use the formula for the number of times a prime \( p \) divides \( n! \):
\[ \text{Number of factors of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
Here, \( p = 11 \).

We need:
\[ \left\lfloor \frac{n}{11} \right\rfloor + \left\lfloor \frac{n}{11^2} \right\rfloor \geq 2 \]

### Step 3: Calculate for different \( n \)
- For \( n = 11 \):
\[ \left\lfloor \frac{11}{11} \right\rfloor + \left\lfloor \frac{11}{121} \right\rfloor = 1 + 0 = 1 \quad (\text{not enough}) \]
- For \( n = 22 \):
\[ \left\lfloor \frac{22}{11} \right\rfloor + \left\lfloor \frac{22}{121} \right\rfloor = 2 + 0 = 2 \quad (\text{sufficient}) \]

### Step 4: Finding the minimum \( n \) for \( 2^{11} \)
Next, we need to ensure that \( n! \) contains at least 11 factors of 2:
\[ \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{8} \right\rfloor + \left\lfloor \frac{n}{16} \right\rfloor \geq 11 \]

- For \( n = 22 \):
\[ \left\lfloor \frac{22}{2} \right\rfloor + \left\lfloor \frac{22}{4} \right\rfloor + \left\lfloor \frac{22}{8} \right\rfloor + \left\lfloor \frac{22}{16} \right\rfloor \]
\[ = 11 + 5 + 2 + 1 = 19 \quad (\text{sufficient}) \]

### Step 5: Conclusion
Since both conditions are satisfied with \( n = 22 \), the least possible value of \( n \) such that \( n! \) is divisible by \( 2^{11} \times 11^2 \) is:

\[ \boxed{22} \]