Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

For every positive integer n, the function h(n) is defined t

Expert replies
by imane81 » Wed Feb 17, 2010 12:04 pm
Thank you for your input on this one

For every positive 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
A. Between 2 and 10
B.Between 10 and 20
C.Between 20 and 30
D.Between 30 and 40
E. Greater than 40
Join the discussion
Source: — Problem Solving |

by komal » Wed Feb 17, 2010 12:15 pm
imane81 wrote:Thank you for your input on this one

For every positive 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
A. Between 2 and 10
B.Between 10 and 20
C.Between 20 and 30
D.Between 30 and 40
E. Greater than 40
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
Join the discussion

by harsh.champ » Thu Feb 18, 2010 3:01 pm
komal wrote:
imane81 wrote:Thank you for your input on this one

For every positive 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
A. Between 2 and 10
B.Between 10 and 20
C.Between 20 and 30
D.Between 30 and 40
E. Greater than 40
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
Hey komal,
I didn't get this :-2*50!+1 will not be divisible by any no <50
Even if we take 50 or 51 how can we confidently say that 2*50!+1 will be divisible by it.The remainder 1 will be there then also.
Kindly clarify.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Join the discussion

by Stuart@KaplanGMAT » Thu Feb 18, 2010 7:00 pm
imane81 wrote:Thank you for your input on this one

For every positive 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
A. Between 2 and 10
B.Between 10 and 20
C.Between 20 and 30
D.Between 30 and 40
E. Greater than 40
This is probably the most asked question on this site (or close to it); do a search on "h(100)" and you'll see lots of explanations.
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
Join the discussion

by vicksikand » Sun Oct 17, 2010 10:30 am
harsh.champ wrote:
komal wrote:
imane81 wrote:Thank you for your input on this one

For every positive 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
A. Between 2 and 10
B.Between 10 and 20
C.Between 20 and 30
D.Between 30 and 40
E. Greater than 40
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
Hey komal,
I didn't get this :-2*50!+1 will not be divisible by any no <50
Even if we take 50 or 51 how can we confidently say that 2*50!+1 will be divisible by it.The remainder 1 will be there then also.
Kindly clarify.
Just FYI:
h(100) = 2^50.50! and not 2.50!
Join the discussion