NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 07:57 It is currently 26 Apr 2024, 07:57
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

For every positive even integer n, the function h(n) is defined to be

SORT BY:
Date
Kudos
   1   2   3 
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
CrackverbalGMAT
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4946
Own Kudos [?]: 7628 [1]
Given Kudos: 215
Location: India
Send PM
Re: For every positive even integer n, the function h(n) is defined to be [#permalink] New post   07 Oct 2021, 22:02
1
Kudos
Top Contributor
Any two consecutive integers will not have any common factor other than 1. We will using this concept to solve this question.
​
\(h(100)\) =\( 2* 4 * 6 *......*100\) = \(2^{50} (1*2*3*4....*50)\) = \(2 ^ {50} * 50!\)

h(100) will be divisible by all prime numbers below 50.
​
Since h(100) and h(100)+1 are consecutive integers and they will not have any common factor other than 1.

We can conclude that smallest prime factor of h(100)+1 will be greater than 50.
​
Option E is the correct answer.

Thanks,
Clifin J Francis,
GMAT SME
​
User avatar
G
BLTN
Senior Manager
Senior Manager
Joined: 25 Aug 2020
Posts: 252
Own Kudos [?]: 116 [0]
Given Kudos: 218
Send PM
Re: For every positive even integer n, the function h(n) is defined to be [#permalink] New post   19 Dec 2021, 23:53
enigma123 wrote:
For every positive even 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

Show Spoiler
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.

h(100) = 2 * 4 * 6 ****************100

Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.

100 = 2 + (n-1) 2
n = 50

Product of terms = Average * number of terms

Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.

H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10


Dear avigutman
Could you validate my approach?

For every positive even 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?

Thus, h(n) = 2*4*6 .... = (2*n+2)!
h(100)+1 = (2*100+2)!+1 = 202!+1
Because 2 consecutive numbers do not share common factors but 1, the answer will be grate then XXX+1 or grater then 40 ?

Thanks beforehand.
User avatar
D
avigutman
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 2287 [1]
Given Kudos: 66
Location: Canada
Schools: NYU Stern (WA)
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Send PM
Re: For every positive even integer n, the function h(n) is defined to be [#permalink] New post   20 Dec 2021, 05:36
1
Kudos
Expert Reply
BLTN wrote:
h(n) = 2*4*6 .... *n = (2*n+2)!


BLTN there's a problem with your logic above.
Just to illustrate the issue with a less abstract situation, consider the following (using your equation with n=4):
h(4) = 2*4 = (2*4+2)! = 10!
But, 2*4 is not equal to 10!

If you want to use factorial for h(n), you'll have to grab a factor of 2 out of every one of the factors of h(n).
For example, if n=10:
h(10) = 2*4*6*8*10 = (2*1) * (2*2) * (2*3) * (2*4) * (2*5) = 2*2*2*2*2 * (1*2*3*4*5) = 2^5 * 5!
User avatar
B
arbazfatmi1994
Manager
Manager
Joined: 05 Jul 2022
Posts: 108
Own Kudos [?]: 13 [0]
Given Kudos: 31
Location: India
WE:Advertising (Health Care)
Send PM
Reviews Badge
Re: For every positive even integer n, the function h(n) is defined to be [#permalink] New post   12 Aug 2023, 07:40
This is a naughty question. :shh: :dazed

Let's go step by step...

h(100) = (2.4.6.8......100)

There are 50 even numbers between 2 and 100 inclusive
Hence there are 50 "2s"

H(100) = 2^50 * (1*2*3*4*......*50)

and (1*2*3*4*......*50) = 50!

So => H(100) = 50!*2^50

H(100) + 1 and H(100) are consecutive numbers.

Note - two consecutive numbers are co-primes
(i.e. they don't have any common factor other than 1)

We can see that H(100) can have any number smaller than 50 as it's factor

So, H(100) + 1 can't have a factor smaller than 50

Hence choose E
GMAT Club Bot
gmatclubot
Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
12 Aug 2023, 07:40
   1   2   3 
Moderators:
Bunuel
Math Expert
92933 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions