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

10
11
12
13
14

990 =9*10*11
so it should contain 9 ,10 and 11
so smallest possible value of n could be 11
(as you can see if you take product from 1 to 11 it will contain 9 ,10,11)

amising6 wrote:If n is a positive integer and the product of all the itnegers from 1 to n, inclusive is a multiple of 990, what is the least possible value of n?

10
11
12
13
14

990 =9*10*11
so it should contain 9 ,10 and 11
so smallest possible value of n could be 11
(as you can see if you take product from 1 to 11 it will contain 9 ,10,11)

Ok Amising,

The factors I found for 990 were 5,2,3,3,11, but when I read n is multiple of 990, I thought n should be greater than 990 (like 15 is multiple of 5), and I see you considered this statement as n is a factor of 990 ... am I wrong with concepts?

Thanks,
Silvia

if you take the product of numjbers from 1 to 11
you will get a number much more than 990

product from 1to 11 will be 39916800 which is a multiple of 990

The question asks for the LEAST possible value of n. Yes, 15 is a multiple of 5, but we would be more interested in 5 being a multiple of 5 in these circumstances.

In this question, the greatest prime factor of 990 is 11, so in order for n to be a multiple of 990, n would have to contain 11 in its prime factorization. Thus, the smallest possible value of n is 11.

True guys,
now that i read again, its the prod of 1..n multiple of 990 ...
And it ask for the least possible value of n ...
I already had the answer, and I didn't see it.
Thank u both
Silvia