The least possible values of n is:
11
We know that the product of first n natural numbers is: n!
Now, we are asked to find the minimum value of n such that n! is a multiple of 990.
This means that:
[tex]n!=990k[/tex]
Now we know that:
[tex]990=2\times 3\times 3\times 5\times 11[/tex]
Now we know that we obtain all of these factors when n is minimum 11.
Since,
[tex]11!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11[/tex]
This means that :
[tex]n\geq 11[/tex]