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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Practice test question
This topic has expert replies
-
scoobydooby
- Legendary Member
- Posts: 1035
- Joined: Wed Aug 27, 2008 10:56 pm
- Thanked: 104 times
- Followed by:1 members
n*(n-1)*(n-2)*.......1 =n!
=>n!=990k (k an integer)
=>n!=9*2*5*11 (for the smallest n, lets take k=1)
if n=10, 10!=10*9*8..1 (we cant get 990, which has a 11)
so the least value of n could be 11, so that 11! would give the 11 we need to make a 990 atleast.
n=11
=>n!=990k (k an integer)
=>n!=9*2*5*11 (for the smallest n, lets take k=1)
if n=10, 10!=10*9*8..1 (we cant get 990, which has a 11)
so the least value of n could be 11, so that 11! would give the 11 we need to make a 990 atleast.
n=11
-
Mr2Bits
- Senior | Next Rank: 100 Posts
- Posts: 73
- Joined: Wed Jan 07, 2009 1:00 pm
- Location: Tampa, FL
- Thanked: 9 times
- GMAT Score:630
When you see the following : product of all the integers from 1 to nBaldini wrote:OA is B. Can you explain how you got the answer?
thanks
Think PRIME numbers. Of the numbers listed, 11 is the smallest prime number that is a multiple of 990
