[AWR83]

kindly assist in solving this question which appeared in one of the GMAT exams:

if n is a positive integer and the product of all the integers from 1 to n inclusive is 990, what is the least possible value of n?

Thanks,

[Capuchin]

I don't quite understand the question. It seems to be asking for the integer n for which n! = 990, but this isn't true for any integer. I also don't understand how you oculd find the "least possible integer" for such a criteria.

[AWR83]

Apologize for this error, the question should read: is a multiple of 990, not equals 990.

Thanks Again.

[galactus]

That should read "The integers from 1 to n, inclusive, is a multiple of 990". The poster left that crucial part out.

Note that 11!/990=8! an integer. The least possible value of n is 11.

[ebaines]

I realize this is an old posting, but I'll add my two cents. The easy way to solve this is to consider the prime factors of 990:

990 = 2*3*3*5*11

The only way for N! To be a multiple of 990 is for it to also be a multiple of 2, 5, 9, and 11. So the smallest possible value for N is 11.