Answer:
Option B.
Step-by-step explanation:
n is a positive integer and product of all integers from 1 to n is a multiple of 990.
That means n! = 990 × k
Or n! = 2 × 5 × 3 × 3 × 11 × k
If n = 2, then 2! will not be a multiple of 990.
If n = 3, then 3! will not be a multiple of 990.
If n = 5, then 5! will not be a multiple of 990.
If n = 9, then 9! will not be a multiple of 990 because 11 will be missing.
Now if we take n = 11, then 11! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 = 39916800 which is divisible by 990.
[As we can see all multiples of 990 = 2, 5, 9, 11 are present in 11! ]
Therefore, option B. 11 will be the answer.
