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?
A) 10
B) 11
C) 12
D) 13
E) 14
Explanation as well, thank you. will post answer soon
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Multiples
This topic has expert replies
-
DanaJ
- Site Admin
- Posts: 2567
- Joined: Thu Jan 01, 2009 10:05 am
- Thanked: 712 times
- Followed by:550 members
- GMAT Score:770
Say P = product of numbers 1 to n. First thing you gotta do is break 990 down into factors: 990 = 99 * 10 = 2*(3^2)*5*11. Now since we have that annoying 11 over there (prime number), it can only mean that n is greater than or equal to 11. So the smallest value of P will be 1*2*...*11, with n = 11. Answer B.
I do not know whether I am reading the question right or not.
The Question says Product of all numbers from 1 to N is a mutliple of 990.
However if you take 11 as the answer
1*2*3*4*5*6*&*8*10*11= 39916800..... which is not a multiple of 990, rather the other way is right.
What Am i missing???
The Question says Product of all numbers from 1 to N is a mutliple of 990.
However if you take 11 as the answer
1*2*3*4*5*6*&*8*10*11= 39916800..... which is not a multiple of 990, rather the other way is right.
What Am i missing???
Sure it is. 990 * 40320 = 11! , i.e. 990n = 11! where n=40320.However if you take 11 as the answer
1*2*3*4*5*6*&*8*10*11= 39916800..... which is not a multiple of 990, rather the other way is right.
What Am i missing???
