DUDE you have made the same mistake while re posting the question.
Here is the actual question.
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n ?
If I take you answer which is 10
Then, what the question stem means is 10! is a multiple of 990.
10! = 10*9*8*7*6*5*4*3*2*1=3628800
990 = 9*10*11
I'll rephrase the question for you.
Is 3628800 a multiple of 990?
The answer is no if n = 10, simply because n! which is 10! does not have 11
Now lets look at 11
11! = 11*10*9*8*7*6*5*4*3*2*1
990 =9*10*11
In this case answer is yes 11! is a multiple of 990.
In the test all you have to do is look at the prime factors.
if x is a multiple of y, then x should have all the prime factors of y, but it is not necessary for y to have all the prime factors of x.
I hope this clears everything. Let me know if you still have any doubts.
All the Best.
