GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 Sep 2018, 02:42

### 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

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

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

Author Message
TAGS:

### Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 505
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

Updated on: 18 Oct 2016, 03:25
56
1
302
00:00

Difficulty:

65% (hard)

Question Stats:

61% (00:47) correct 39% (01:14) wrong based on 2145 sessions

### HideShow timer Statistics

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 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40

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

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Originally posted by enigma123 on 28 Jan 2012, 17:46.
Last edited by Bunuel on 18 Oct 2016, 03:25, edited 4 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 49381
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

28 Jan 2012, 17:53
278
1
248
enigma123 wrote:
h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40
< 30
> 40
< 10
Indeterminate

Below is the proper version of this question:

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 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all prime numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY prime factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest prime factor of $$h(100)+1$$ will be more than 50.

_________________
Intern
Joined: 28 Aug 2012
Posts: 44
Location: Austria
GMAT 1: 770 Q51 V42
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

02 Sep 2012, 11:31
16
2
I'm not taking credit for the following solution. Found it in another forum.

Two consecutive integers cannot be divisible by the same integer greater than 1.

So if we can prove that h(100) is divisible by every number smaller than 50, we proved that h(100)+1 is NOT divisible by any number smaller than 50 (besides 1).

2*4*6*...*98*100=2*(1*2*3*...*50)=2*(50!)
Hence, h(100) is divisible by every number smaller than 51. So the smallest prime factor of h(100)+1 is at least 53.
##### General Discussion
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 505
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

28 Jan 2012, 18:00
3
1
Bunuel - how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 49381
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

28 Jan 2012, 18:08
51
14
enigma123 wrote:
Bunuel - how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain.

Given that the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive, so: $$h(100)=2*4*6*...*100=(2*1)*(2*2)*(2*3)*(2*4)*...*(2*50)$$ --> factor out all 50 2's: $$h(100)=2^{50}*(1*2*3*..*50)=2^{50}*50!$$.

Hope it's clear.
_________________
Intern
Joined: 30 Nov 2011
Posts: 29
Location: United States
GMAT 1: 700 Q47 V38
GPA: 3.54
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

02 Mar 2012, 13:26
2
Thanks Bunuel. It was very helpful. This was my 3rd question on the practice test, I spent a minute without even knowing where to start from, so I made a random guess and moved on. I got 13 incorrect questions out of the 37, but I managed to score 48. thanks again
Intern
Joined: 04 Apr 2011
Posts: 43
Location: India
Concentration: Marketing, Entrepreneurship
GPA: 3.8
WE: Marketing (Real Estate)
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

01 May 2012, 06:42
1
Thanks for the explanation. I also got this as my 3rd question, couldn't figure it out, ended up guessing and moving on. Surprising that GMATPrep throws you such hard questions early.
Intern
Joined: 28 Apr 2012
Posts: 21
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

10 May 2012, 06:53
2
1
Bunuel wrote:
enigma123 wrote:
h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40
< 30
> 40
< 10
Indeterminate

Below is the proper version of this question:
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 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all prime numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY prime factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest prime factor of $$h(100)+1$$ will be more than 50.

That is an impressive solution. <-- understatement
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

10 May 2012, 07:08
26
9
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 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

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

Check out this post for detailed theory involved in this question: http://www.veritasprep.com/blog/2011/09 ... h-part-ii/
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Director
Affiliations: GMATQuantum
Joined: 19 Apr 2009
Posts: 605
For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

Updated on: 17 Sep 2018, 00:44
15
20
Stiv,

Here are some other official GMAT questions that are conceptually related:

1) GMAT Prep: http://gmatclub.com/forum/does-the-inte ... 26735.html

2) Official Guide GMAT 13th Edition: http://www.beatthegmat.com/divisibility-t111432.html
Video explanation: https://gmatquantum.com/official-guides ... h-edition/

3) Video explanation to the original problem(For every positive even integer n, the function h(n) is....): https://gmatquantum.com/gmatprep-number ... n-defined/

Dabral
Attachments

image12.png [ 33.64 KiB | Viewed 189244 times ]

Originally posted by dabral on 12 May 2012, 02:45.
Last edited by dabral on 17 Sep 2018, 00:44, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

12 May 2012, 03:10
3
2
Stiv wrote:

Karishma, could you recommend me some other on-line material or sites where I can study these kind of problems in-depth? Thank you!

Check out part i of the link given above as well. That should cover the theory that is useful for such questions. As for examples, you can search for 'factors consecutive integers' in the Quant forum and you should hit quite a few questions based on these concepts.
I got one in one of my posts: if-n-is-a-positive-integer-and-r-is-the-remainder-when-119518.html?hilit=factor%20consecutive%20integers
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 27 Feb 2012
Posts: 37
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

29 May 2012, 20:30
hifunda wrote:
Thanks for the explanation. I also got this as my 3rd question, couldn't figure it out, ended up guessing and moving on. Surprising that GMATPrep throws you such hard questions early.

this was my second question on the first practice test, i'll have to review this answer quite a few times until it sticks. Great answer
Manager
Joined: 03 Jul 2012
Posts: 112
GMAT 1: 710 Q50 V36
GPA: 3.9
WE: Programming (Computer Software)
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

27 Aug 2012, 05:07
1
Hey Bunuel, I used the following logic, and got E.
(2*4*6*8.... 100) +1 is of the format 6p +1. Therefore, its a prime number and has only two factors, one and itself. So, its definitely greater than 40.

So, E

Is it correct?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

27 Aug 2012, 22:52
5
3
mehulsayani wrote:
Hey Bunuel, I used the following logic, and got E.
(2*4*6*8.... 100) +1 is of the format 6p +1. Therefore, its a prime number and has only two factors, one and itself. So, its definitely greater than 40.

So, E

Is it correct?

Every prime number greater than 3 is of one of the two forms: (6a -1), (6a + 1)
e.g. 5 = 6*1 - 1; 7 = 6*1 + 1; 11 = 6*2 - 1; 13 = 6*2 + 1 etc

But every number of the form (6a -1) or (6a + 1) is not prime.
e.g. 25 = 6*4 + 1 i.e. it is of the form 6a+1. But 25 is not prime.

So just because a number is of the form 6a + 1, we cannot say that it is definitely prime.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Joined: 14 Jun 2012
Posts: 54
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

21 Sep 2012, 01:49
3
Got this one on the 9th question of my GMATPrep test. I actually reached a good point on this problem.

I got to the point that the product of 2*4*6*....*100 = (2^50)*50!

At this point I lost the plot and developed a weird approach here.

Since the last number in the given h(n) was 100 so the actual value could be like xxx00. So now, h(n) + 1 would be some value like xxx01.

So the possible prime factors are effectively 3, 7, 11, 13 etc.

I simply gambled on one of these till 20 being a factor and chose A. It was wrong though. The correct answer is indeed E.

The explanations above specially the one by Bunuel where the point of co-prime numbers not having any common factors but 1 was particularly good stuff and I had not read it before. This information will surely help the next time provided I am able to fully recognize it down.
_________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

There are no shortcuts to any place worth going.

VP
Joined: 24 Jul 2011
Posts: 1486
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

22 Apr 2013, 22:40
2
2
h(n) = the product of all even integers from 2 to n inclusive
=> h(100) + 1 = (2 * 4 * 6 * 8 * ....... * 100) + 1
= 2^50 * (1*2*3*4*......*50) + 1

As all the numbers from 2 to 50 are factors of 2^50 * (1*2*3.....*50), none of these can be factors of 2^50 * (1*2*3*4*......*50) + 1 (i.e. of h(100)+1)

Therefore the smallest prime factor of h(100) + 1 is greater than 50.

This corresponds to option (E).
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

22 Apr 2013, 22:45
3
2
itsmeabhi99 wrote:
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

I have discussed this question and the related theory in detail here: http://www.veritasprep.com/blog/2011/09 ... h-part-ii/
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Joined: 26 Jan 2013
Posts: 270
Schools: Stanford '19
GMAT 1: 770 Q51 V44
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

12 May 2013, 12:19
2
1
This is a great question. Let me try to post a simple solution!

Let's divide the solution into two parts:

Part (1):

We need to simplify h(100) first.

h(100) = 2*4*6*8*....*98*100.

Consider that 2 = 2*1, 4 = 2*2, etc. up to 100 = 2*50.

h(100) then simplifies to $$h(100) = 2*1 * 2*2 * 2*3 *.... * 2*49 * 2*50 = 2^{50} * 1*2*3*4*... *50$$ (A)

Part (2):

Consider the second part of the question. We need to find a bound for the smallest prime factor of h(100)+1. Any number p that is a factor of h(100)+1, leaves a remainder of (p-1) when it divides h(100).

remainder(h(100)/p) = p-1.

Another way of saying this is "p is a prime number that is not a divisor of h(100)" (B)

From (A):

Clearly h(100) is divisible by all prime numbers less than 50 (as h(100) is divisible by 50!). The smallest prime that is not a divisor of h(100) should therefore be greater than 50.

Looking at the choices, the most appropriate choice is:

(e) greater than 40.

Hope that Helps!
_________________

Karthik P.
Founder and CEO,

We are an India based MBA admissions consulting firm. We've been helping Indian MBA applicants (students belonging to the most competitive MBA admissions pool) since 2013. Several of our students study at top 20 schools in the US including Kellogg, MIT, Yale, Ross and Darden, top European schools including London Business School and Oxford, top Canadian schools like Rotman and top Asian schools like ISB and NUS. Our students have won over USD 2.1 Million in scholarships over the last 3 admissions years.

Schedule a free consultation with us.

If you like this post - please give us kudos Thanks much!

Math Expert
Joined: 02 Sep 2009
Posts: 49381
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

12 Jun 2013, 05:25
1
1
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Number Properties: math-number-theory-88376.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

24 Mar 2014, 03:12
Responding to a pm:

Quote:
The question here talks about h(50)+1

Last term number n=25
Average = 26
Sum = 26*25=650

h(50)+1 = 651 which is not a prime number as it is divisible by 3. Please correct me if i m wrong here. my answer is A.

I am not sure where you got this question since the original question here talks about h(100) + 1. Even if we do assume h(50) + 1, note that h(2) = 2
h(4) = 2*4
h(6) = 2*4*6
....
So I am not sure how you got average 26.
The correct answer here is (E) as explained in the link given in my post above.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Re: For every positive even integer n, the function h(n) is defined to be &nbs [#permalink] 24 Mar 2014, 03:12

Go to page    1   2   3    Next  [ 54 posts ]

Display posts from previous: Sort by

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

## Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.