A tough one - h(100) + 1

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Sat Feb 20, 2010 11:22 am

A tough one - h(100) + 1

by papumba2011 » Mon Apr 12, 2010 12:30 pm
For every postive integer n , the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is
1) between 2 and 10
2) between 10 and 20
3) between 20 and 30
4) between 30 and 40
5) greater than 40
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 362
Joined: Fri Oct 02, 2009 4:18 am
Thanked: 26 times
Followed by:1 members

by indiantiger » Mon Apr 12, 2010 12:49 pm
I think this question has been answered on the forum, just search for h(n) and you will get multiple hits.

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Mon Apr 12, 2010 1:00 pm
indiantiger wrote:I think this question has been answered on the forum, just search for h(n) and you will get multiple hits.
It's been answered h(n) + 1 times at least!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Senior | Next Rank: 100 Posts
Posts: 65
Joined: Wed Nov 25, 2009 6:33 pm
Thanked: 3 times

by dxgamez » Mon Apr 12, 2010 10:57 pm
lol that's a good one, Stuart...

@papumba

the number properties which is important to note in this qn is that consecutive integers do not share any primes. that would mean that h(100) and h(100) + 1 do not share any primes. you can get the solutions from the forum.

thought i should write it down, to remind myself too :)

Legendary Member
Posts: 809
Joined: Wed Mar 24, 2010 10:10 pm
Thanked: 50 times
Followed by:4 members

by akhpad » Tue Apr 13, 2010 4:03 am
This problem has been answered many times.

Please type first few words in google. You will get it at several places.