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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
zaur, as a rule, if n is divisible by a factor, (n+1) will not be divisible by the same factor. You can check this for any number. For example, if 6 is divisible by 2, 3 and 6 then (6+1) will not be divisible by any of 2,3, or 6.
Extending the same logic, h(100) is divisible by all the integers from 1 to 50. Hence, h(100)+1 will not be divisible by any number from 1 to 50 and hence its factor will be greater than 50.
Thanks scthakur! Its much clear now! + 1 kudo for you.
For every positive 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
a. between 2 an 10 b. between 10 and 20 c. between 20 and 30 d. between 30 and 40 e. greater than 40
Thanks - I am sure its easy I just don't know how to attack it.
Re: GMATPrep Question [#permalink]
25 Nov 2008, 20:46
3
This post received KUDOS
you can se that :h(100)=2*4*6*...*100=(2*1)(2*2)(2*3)...(2*50) or h(100)=2^50(1*2*3...*50) h(100)+1=2^50(1*2*3*...*50) + 1 if h(100)+1 is divisible by p then p must be greater than 47. Here we go: we can see that h(100) is divisible by 2,3,..47(47 is the greatest prime which is smaller than 50) so if h(100) +1 is is divisible by p which p <=47 then 1 is divisible by p so p=1 ridiculous then if h(100) +1 divisible by p, p must be greater than 47 and answer is E We use the characteristic : a+ b is divisible by c, and a is divisible by c too. So b is divisible by c
Re: GMATPrep Question [#permalink]
26 Nov 2008, 17:34
So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.
Re: GMATPrep Question [#permalink]
27 Nov 2008, 01:32
bertlacy wrote:
So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.
i am one of those slightly intel.
however this is based on a concept that says
if x is devisible by z then sure x+1 is not devisible by z also, the largest prime that devide x has to be less than the largest prime that devide x+1.
This is probably the most hardest question i ever encountered.
h (2) = 2 h (4) = 2 x 4 = 2! x 2^2 h (6) = 2 x 4 x 6 = 3! x 2^3 h (8) = 2 x 4 x 6 x 8 = 4! x 2^4 h (10) = 2 x 4 x 6 x 8 x 10 = 5! x 2^5 . . . . h (100) = 2 x 4 x 6 x .........x 100 = 50! x 2^50 = (1x2x3x4x5x......x40x....48x49x50) (2^50). The pattren identified is that for every n, h (n) = (n/2)! x 2^(n/2)
Now lets find the p, the smallest prime factor of h (100) + 1. lets try from the begaining:
if p = 2, [h(100)+1] should be divisible by 2. But it is not. [h (100) + 1]/2 = h (100)/2 + 1/2 = (50! x 2^50)/2 + 1/2. it doesnot result in an integer. so 2 cannot be p.
if p = 3, [h (100) + 1]/3 = h (100)/3 + 1/3 = (50! x 2^50)/3 + 1/3. it doesnot result in an integer. so 3 cannot be p. if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 3 cannot be p. if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 4 cannot be p. if p = 5, [h (100) + 1]/5 = h (100)/5 + 1/5 = (50! x 2^50)/5 + 1/5. it doesnot result in an integer. so 5 cannot be p. . . . . . if p = 50, [h (100) + 1]/50 = h (100)/50 + 1/50 = (50! x 2^50)/50 + 1/50. it doesnot result in an integer. so 50 cannot be p.
The finding is that: Something [h(n)] divisible by k is not divisible by k if 1 is added to something [h(n)]. Therefore h (n=100) + 1 has the smallest prime i.e. > 50. So it is E. _________________
PS -- smallest prime factor [#permalink]
22 Feb 2009, 11:18
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 100 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40.
please explain ... this question is from gmatprep1.
Smallest Prime Factor [#permalink]
14 Mar 2009, 15:26
For every positive 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 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40
Schools: HBS(08) - Ding. HBS, Stanford, Kellogg, Tuck, Stern, all dings. Yale - Withdrew App. Emory Executive -- Accepted, Matriculated, Withdrewed (yes, I spelled it wrong on purpose). ROSS -- GO BLUE 2011.
Re: Smallest Prime Factor [#permalink]
16 Mar 2009, 11:30
nitindas wrote:
For every positive 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 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40
Re: Smallest Prime Factor [#permalink]
17 Mar 2009, 13:36
RahlowJenkins wrote:
nitindas wrote:
For every positive 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 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40
the smallest prime factor of 2*4*6*8...*100 is 2.
2+1=3
answer is 1.
I gues this is not correct . You did not take the +1 in consideration , which changes the answer. Please see my response below. _________________
Lahoosaher
gmatclubot
Re: Smallest Prime Factor
[#permalink]
17 Mar 2009, 13:36
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...