Author 
Message 
TAGS:

Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 505
Location: United Kingdom
Concentration: International Business, Strategy
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
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 + (n1) 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 + (n1) 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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
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 coprime, 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. Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 28 Aug 2012
Posts: 44
Location: Austria

Re: For every positive even integer n, the function h(n) is defined to be
[#permalink]
Show Tags
02 Sep 2012, 11:31
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.




Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 505
Location: United Kingdom
Concentration: International Business, Strategy
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
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



Intern
Joined: 30 Nov 2011
Posts: 29
Location: United States
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
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
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
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 coprime, 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. Answer: E. 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
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 + (n1) 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 + (n1) 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 ... hpartii/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy 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
Stiv, Here are some other official GMAT questions that are conceptually related: 1) GMAT Prep: http://gmatclub.com/forum/doestheinte ... 26735.html2) Official Guide GMAT 13th Edition: http://www.beatthegmat.com/divisibilityt111432.htmlVideo explanation: https://gmatquantum.com/officialguides ... hedition/3) Video explanation to the original problem(For every positive even integer n, the function h(n) is....): https://gmatquantum.com/gmatprepnumber ... ndefined/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
Stiv wrote:
Karishma, could you recommend me some other online material or sites where I can study these kind of problems indepth? 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: ifnisapositiveintegerandristheremainderwhen119518.html?hilit=factor%20consecutive%20integers
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy 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
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
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
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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy 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
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 coprime 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 BSchool Journey in a Blog : tranquilnomadgmat.blogspot.com
There are no shortcuts to any place worth going.



VP
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1486

Re: For every positive even integer n, the function h(n) is defined to be
[#permalink]
Show Tags
22 Apr 2013, 22:40
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
Top MBA Admissions Consulting  Top MiM Admissions Consulting
Premium MBA Essay ReviewBest MBA Interview PreparationExclusive 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
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 ... hpartii/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



MBA Admissions Consultant
Joined: 26 Jan 2013
Posts: 270

Re: For every positive even integer n, the function h(n) is defined to be
[#permalink]
Show Tags
12 May 2013, 12:19
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 (p1) when it divides h(100). remainder(h(100)/p) = p1. 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, August Academy.
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.
Click here to view GMATclub verified student testimonials.
Click here to view all our student testimonials.
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



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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy 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 ]



