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 →

Functions

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Wed Sep 01, 2010 10:04 amFollowed by: 20 members
Thanked: 1 times
by Akansha » Thu May 26, 2011 8:59 am
For every positive even integer n, the function h(n) is defined to be the
product of all even integers from 2 to n inclusive. If p is the smallest prime factor
of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members
Thanked: 72 times

by manpsingh87 » Thu May 26, 2011 9:07 am
Akansha wrote:For every positive even integer n, the function h(n) is defined to be the
product of all even integers from 2 to n inclusive. If p is the smallest prime factor
of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E
h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members
Location: India
Thanked: 310 times
GMAT Score:750

by cans » Fri May 27, 2011 4:20 am
manpsingh87 wrote:
h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E
instead of 47, smallest prime factor must be greater than 50.. right??
Answer will remain the same..
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Fri Jan 14, 2011 10:01 am
Thanked: 10 times

by djiddish98 » Fri May 27, 2011 4:33 am
It's basically the same thing since there aren't any prime factors between 47 and 50.

mapsingh did a nice job of explaining it.

One way to look at it is through the divisibility rules.

If X is divisible by Y and and Z is NOT divisible by Y, then X+Z will NOT be divisible by Y.

Based on mapsingh's analysis, we know that our (h100) term is divisible by all possible primes up to 50 (or 47). 1 is not divisible by any prime number, therefore our h(100)+1 term will not be divisible by any term that (h100) is divisible by.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri May 27, 2011 6:15 am
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
Options:
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers. Consecutive integers are co-primes: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common both to x and to x+1. Integers that share no factors other than 1 are called coprimes.

Thus, in the problem above, h(100) and h(100)+1 are coprimes. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100

Factoring out 2, we get:

h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). Since h(100) and h(100)+1 share no factors other than 1, none of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.
Last edited by GMATGuruNY on Sun Jan 20, 2013 11:44 am, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 226
Joined: Thu Nov 25, 2010 12:19 amFollowed by: 2 members
Thanked: 3 times

by nafiul9090 » Mon May 30, 2011 5:18 pm
manpsingh87 wrote:
Akansha wrote:For every positive even integer n, the function h(n) is defined to be the
product of all even integers from 2 to n inclusive. If p is the smallest prime factor
of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E
h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E
hello

could you please elaborate it further, how do i reach it h(100)+1= 2^50(1*2*3*4*5.....50)+1??

i cant get it

regards nafi

thanks in advance
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members
Thanked: 72 times

by manpsingh87 » Mon May 30, 2011 7:36 pm
nafiul9090 wrote:
manpsingh87 wrote:
Akansha wrote:For every positive even integer n, the function h(n) is defined to be the
product of all even integers from 2 to n inclusive. If p is the smallest prime factor
of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E
h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E
hello

could you please elaborate it further, how do i reach it h(100)+1= 2^50(1*2*3*4*5.....50)+1??

i cant get it

regards nafi

thanks in advance
hi, the function h(n) is defined to be the product of all even integers from 2 to n inclusive;

so, h(10)= 2*4*6*8*10;
=2*1*2*2*2*3*2*4*2*5;
= 2^5(1*2*3*4*5);

similarly h(100)=2*4*6*......100;
=2^50(1*2*3*4*5....50);

i hope it helps..!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Mon Dec 24, 2012 5:49 pm

by Theo » Mon Dec 24, 2012 6:05 pm
Cheers for the method!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1