What is the smallest natural number n such that `n!` is divisible by 990?

Options

A. 9
B. 11
C. 33
D. 99

Answer

To find the smallest natural number \( n \) such that \( n! \) is divisible by 990, we can follow these steps:

### Step 1: Factorize 990
First, we need to factorize 990 into its prime factors.
\[ 990 = 9 \times 110 = 9 \times (11 \times 10) = 9 \times 11 \times (2 \times 5) = 11 \times 9 \times 5 = 11 \times 3^2 \times 5 \]
Thus, the prime factorization of 990 is:
\[ 990 = 2^1 \times 3^2 \times 5^1 \times 11^1 \]

### Step 2: Determine the smallest \( n \)
Next, we need to find the smallest \( n \) such that \( n! \) contains at least these prime factors in the required quantities.

- **For \( 2^1 \)**: The smallest \( n \) that includes at least one factor of 2 is \( n = 2 \).
- **For \( 3^2 \)**: The smallest \( n \) that includes at least two factors of 3 is \( n = 6 \) (since \( 6! = 720 \) which includes \( 3 \times 2 \times 1 \)).
- **For \( 5^1 \)**: The smallest \( n \) that includes at least one factor of 5 is \( n = 5 \).
- **For \( 11^1 \)**: The smallest \( n \) that includes at least one factor of 11 is \( n = 11 \).

### Step 3: Find the maximum \( n \)
Now we take the maximum of these values to ensure \( n! \) contains all the necessary factors:
\[ \text{max}(2, 6, 5, 11) = 11 \]

### Conclusion
Thus, the smallest natural number \( n \) such that \( n! \) is divisible by 990 is:
\[ \boxed{11} \]