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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

gmat prep q..

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 51
Joined: Sat Mar 22, 2008 1:08 am
Thanked: 2 times
by abhaypratapsingh » Sun Jul 20, 2008 6:29 am
h(n) is product of all even integers 2 to n, inclusive. if P is smallest prime factor of h(100) + 1 , the n P is":

between 2 to 10
between 10 to 20
between 20 to 30
between 30 to 40
greater than 40
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 189
Joined: Thu May 01, 2008 10:55 pmFollowed by: 1 member
Location: Seattle, WA
Thanked: 25 times
GMAT Score:750+

by VP_Tatiana » Sun Jul 20, 2008 9:15 am
For this type of problem, it is easiest to solve for the most simple case and then generalize. If they are asking us for a general answer, and there is not an answer choice like "E: None of the above", then we know that no matter what n we pick, P must be in the same range.

So, I pick n = 2 so that h(n) = 2.

h(100) + 1 = 201 = 3*67 (I quickly tell that 201 is divisible by 3 because 2+0+1 = 3)

So, the smallest prime factor of the equation is 3, and the answer is "between 2 and 10."
Tatiana Becker | GMAT Instructor | Veritas Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Sun Jul 20, 2008 2:08 pm
VP_Tatiana wrote:For this type of problem, it is easiest to solve for the most simple case and then generalize. If they are asking us for a general answer, and there is not an answer choice like "E: None of the above", then we know that no matter what n we pick, P must be in the same range.

So, I pick n = 2 so that h(n) = 2.

h(100) + 1 = 201 = 3*67 (I quickly tell that 201 is divisible by 3 because 2+0+1 = 3)

So, the smallest prime factor of the equation is 3, and the answer is "between 2 and 10."
You are not free to pick a value of n here; the question asks specifically about h(100) + 1. That is, n = 100.

What is h(100)?

h(100) = 2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50) = (2^50)*50!

So we need to know about the smallest prime factor of (2^50)*50! + 1.

Notice that 50! is divisible by every prime less than 50. That ensures that (2^50)*50! is divisible by every prime less than 50, which ensures that (2^50)*50! + 1 will be divisible by no prime less than 50: the remainder will be 1 each time. The smallest prime factor of h(100) + 1 must therefore be larger than 50 (and therefore certainly larger than 40). E.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 20
Joined: Tue Mar 25, 2008 7:37 pm

by darcknyht » Mon Jul 21, 2008 6:12 am
Ian,

I spent forever trying to even write that out on my own.

You are the man.

Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Wed Sep 10, 2008 12:16 pmFollowed by: 1 member
Location: Glendale, AZ
Thanked: 3 times

by jaybrium » Tue Oct 28, 2008 10:27 am
Well done, Ian. Thanks!
- Jay
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
MBA Student
Posts: 1194
Joined: Sat Aug 16, 2008 9:42 pmFollowed by: 17 members
Location: Paris, France
Thanked: 71 times
GMAT Score:710

by gmat740 » Sat Jul 18, 2009 7:25 pm
Ian Stewart wrote:
VP_Tatiana wrote:For this type of problem, it is easiest to solve for the most simple case and then generalize. If they are asking us for a general answer, and there is not an answer choice like "E: None of the above", then we know that no matter what n we pick, P must be in the same range.

So, I pick n = 2 so that h(n) = 2.

h(100) + 1 = 201 = 3*67 (I quickly tell that 201 is divisible by 3 because 2+0+1 = 3)

So, the smallest prime factor of the equation is 3, and the answer is "between 2 and 10."
You are not free to pick a value of n here; the question asks specifically about h(100) + 1. That is, n = 100.

What is h(100)?

h(100) = 2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50) = (2^50)*50!

So we need to know about the smallest prime factor of (2^50)*50! + 1.

Notice that 50! is divisible by every prime less than 50. That ensures that (2^50)*50! is divisible by every prime less than 50, which ensures that (2^50)*50! + 1 will be divisible by no prime less than 50: the remainder will be 1 each time. The smallest prime factor of h(100) + 1 must therefore be larger than 50 (and therefore certainly larger than 40). E.
The smallest prime factor if h(100)+1 > 47(although both mean the same. But if in the question, we had two answer choices then I think it would be important to break down the problem into basic terms.Please take a look.

h(100) = 2*4*6 ....... 96*98*100. The maximum possible prime factor for h(100) can be calculated by checking :
98/2 = 49 (not prime), 96/2 = 48(not prime), 94/2 = 47 (prime).
So maximum possible prime factor of h(100) =47.

Since p is a prime factor of h(100)+1, so p should atleast be greater than 47 to be a prime factor of h(100)+1.

Hence p>47

I hope I am clear about my approach.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 157
Joined: Tue Oct 07, 2008 5:47 am
Thanked: 3 times

by PussInBoots » Sat Jul 18, 2009 8:13 pm
Very interesting
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1799
Joined: Wed Dec 24, 2008 3:03 amFollowed by: 2 members
Thanked: 36 times

by goelmohit2002 » Mon Jul 27, 2009 12:05 pm
IMO..we can solve the same even more simply...even without bothering about finding the prime numbers less than 50.....

The property of numbers that we need to use is

"No two consecutive positive integers(n, n+1) are ever divisible by same number except 1."

Now if we see that h(n) =(2^50) * 50!.

so it is divisible by all numbers from 2-50....

so the minimum number(including prime/non prime etc etc..) that will divide h(n) will be 51......

Thus it leads to E.[/i]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1