BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

990

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Mar 26, 2011 4:59 am
Thanked: 4 times

990

by B166418 » Wed Jun 29, 2011 10:50 am
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
Top
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Wed Jun 29, 2011 10:53 am
Hi,
990 = 9*10*11
11 is a prime number so we need n! to be at least 11! to be divisible by 990.

Hence, B
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pm
Location: Arlington, MA.
Thanked: 27 times
Followed by:2 members

by winniethepooh » Wed Jun 29, 2011 12:12 pm
Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Thu Jun 30, 2011 2:28 am
winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Hi,
I think you have misinterpreted the question. What the question says is:
Find the least value of n such that 990 is a factor of n!.
990 = 2*3^2*5*11
Now for 990 to be a factor of n!, 11 should be a factor of n!. The least value of n that satisfies this condition is 11.
Consider 10! = 1.2.3...10
11 is not a factor of this right?
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 461
Joined: Tue May 10, 2011 9:09 am
Location: pune
Thanked: 36 times
Followed by:3 members

by amit2k9 » Thu Jun 30, 2011 2:44 am
990 = 11*10*9
thus 11!.
B it is.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
Top
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pm
Location: Arlington, MA.
Thanked: 27 times
Followed by:2 members

by winniethepooh » Thu Jun 30, 2011 6:32 am
Thanks Frankenstine!
Top
Senior | Next Rank: 100 Posts
Posts: 41
Joined: Mon Jan 10, 2011 4:54 pm
Thanked: 1 times

by Buix0065 » Fri Jul 01, 2011 2:43 pm
Frankenstein wrote:
winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Hi,
I think you have misinterpreted the question. What the question says is:
Find the least value of n such that 990 is a factor of n!.
990 = 2*3^2*5*11
Now for 990 to be a factor of n!, 11 should be a factor of n!. The least value of n that satisfies this condition is 11.
Consider 10! = 1.2.3...10
11 is not a factor of this right?
Hi Frankenstein,

Just to make sure I understand your reasoning.

1. a number is a factor if all primes pair
2. 990 has 2, 5, and 11 that need to pair
3. the smallest that n can be to ensure n! will pair 2, 5, 11 is 11?

Thanks for your insights!
Top
Post new topic Post Reply
• Page 1 of 1