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:

For every positive integer n, the function h(n) is defined [#permalink]

Show Tags

04 Jan 2009, 21:46

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 smalled prime factor of h(100)+1 then p is ?

a) between 2 and 10 b) between 20 and 30 c) between 30 and 40 d) >40

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

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 smalled prime factor of h(100)+1 then p is ?

a) between 2 and 10 b) between 20 and 30 c) between 30 and 40 d) >40

Re: GMATPrep question - to Gmat Tiger [#permalink]

Show Tags

05 Jan 2009, 10:34

Hi Gmat Tiger, I am not saying your explanation in the link is wrong but there is something I don't understand. You state that the number can not be divisible by any factor because according to the formula (2^50 x 50!)/ factor + 1/factor will always result in a non integer number?

Imagine that instead of h(50) we calculate h(8): (2^4 x 4!)/factor + 1/factor. According to your theory whatever is the factor the result will always be not integer. But if you try the factor 5 the result is an integer (385/5 = 77).

I am not sure that I understood your approach and so far I have not solved the problem in any other way, but I would like to know if there is something that I am missing.

Re: GMATPrep question - to Gmat Tiger [#permalink]

Show Tags

05 Jan 2009, 23:55

gmark wrote:

Hi Gmat Tiger, I am not saying your explanation in the link is wrong but there is something I don't understand. You state that the number can not be divisible by any factor because according to the formula (2^50 x 50!)/ factor + 1/factor will always result in a non integer number?

Imagine that instead of h(50) we calculate h(8): (2^4 x 4!)/factor + 1/factor. According to your theory whatever is the factor the result will always be not integer. But if you try the factor 5 the result is an integer (385/5 = 77).

I am not sure that I understood your approach and so far I have not solved the problem in any other way, but I would like to know if there is something that I am missing.

thx

I did not quite understand the red part. but for me you are talking beyond the scope of the question.

[h(8)+1] and [h(100)+1] look similar but they are different issues and the rule may not be applied to h(8)+1. and I even did not say that the rule applied to any function. stick to [h(100)+1].

Take a time and go agin and understand the concept. you will get it as i did so.

In fact, I always tried to skip this question as it was always difficult for me. this time I tried so many minuets (almost 30) to solve it, understanding the concept. _________________

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50.Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

There you go: h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. _________________

I don't know if I am doing this right. Please correct if I am wrong.

product of all even integers ... 2*4*6*... factor 2 out and you have 2^50*(1*2*3...) = (2^50)*50!. so the question is what is the smallest prime factor for (2^50)*50! + 1. Any number less than 50 is always a factor of (2^50)*50! leaving reminder 1. so it has to be greater than 50

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

nice explanation. I think I'd poop myself if I saw a question this hard on the exam. Definitely the hardest one I've encountered

gmatclubot

Re: GMATPrep question: need solution
[#permalink]
08 Feb 2009, 16:50