pratyoosh wrote:Q. 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?

1. 10
2. 11
3. 12
4. 13
5. 14

Ans: B

1 * 2 * 3 * .... * n = n!

Lets find out the prime factors of n! = 990 = 10 * 99 = 2 * 3 * 3 * 5 * 11

So, for our number to be a multiple of 990, it must include those 4 prime factors.

Since the biggest prime on the list is 11, 11! is the smallest factorial that will be a multiple of 990.

The reason we can directly choose so is 11 is prime, hence we are sure that there is no other factors available below 11 to make a product 11. Thus least possible value of 990 is 11.

Basically what the question says is that the product of all integers from 1 to n or n! is a multiple of 990.

In other words : n!/990 must be an integer,for if it isn't n! is not gonna be a multiple of 990.[eg 77 is a multiple of 11 therefore 77/11 yields an integer.similarly 76 is not a multiple of 11;therefore,76/11 doe not yield an integer]

Simplifying : n!/(9*10*11). Now n! must be divisible by each to be an integer.Being divisible by just one is not enough.It needs to be divisible by each to be an integer.

Now we know 11! includes,11,10 and 9.Therefore,the smallest value that n can be is 11.If n=10,then it wont be divisible by 11 (it is a prime number).]

n could be 12,13,or 14.In each case n would yield an integer on being divided by 990.But the question is smallest;therefore,the answer should be 11.