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 even integer n, the function h(n) is defined to be 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 1) between 2 and 10 2) between 10 and 20 3) between 20 and 30 4) between 30 and 40 5) greater than 40

OMG this question is like a ghost that keeps on appearing here every other week... It has been discussed many many times... just google (or search our forum for "smallest prime factor" Just one of the discussions is here... smallest-prime-factor-76565.html _________________

This question is testing the concept of coprimes. 2 positive integers are coprime when their greatest common factor (their only common factor) is 1. Now note that two different prime numbers are always coprime. For example, 3 and 7 are coprime. So are 13 and 19. But the two integers need not be prime numbers in order to be coprime. For example, 4 and 9 are coprime (1 is their only common factor). Also, Important! Two consecutive integers are always coprime. The question is testing you on this concept. Let's solve it now: h(n) = 2*4*6*....................*n. ----n is even. h(100) = 2*4*6*........ 94*96*98*100. h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100). Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47. Clearly, this prime factor is greater than 40.

Spot on GMAT Doctor +1 Kudos ; I had no idea whatsoever... I was prolly gonna guess E anyways.. but knowing the approach, phew.... great relief knowing what u r doing ) _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Pl. provide the solution for the problem? [#permalink]

Show Tags

16 Nov 2009, 13:30

For every +ve even integer n, the function h(n) is defined as the product of all even integer 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) b/w 10 & 20 c) b/w 20 & 30 d) b/w 30 & 40 e) greater than 40

Pl. provide the solution...thanx in advance Correct answer is e

Question is - 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 factor of h(100) + 1, then p is

A. between 2 and 10 B. between 20 and 10 C. between 20 and 30 D. between 30 and 40 E. greater than 40

OA is (e) , but I am not able to figure out why it is (e)

Question is - 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 factor of h(100) + 1, then p is

a) between 2 and 10 b) between 20 and 10 c) between 20 and 30 d) between 30 and 40 e) greater than 40

OA is (e) , but I am not able to figure out why it is (e)

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

Wow, lot of info. I understood everything except for this part: h(100) = (2^50)*(1*2*3*.......47*48*49*50) Could you explain this part a little more clearly? Thanks.

For every positive even integer n, the function h(n) is defnied 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

Any help on how to get the correct answer? Thanks!

What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?

Answer: 16

Since there is a hypotenuse, we are talking of a right isosceles triangle. Let a, b and c be sides of the triangle, where a=b and c is the hypotenuse. from the prompt, a+b+c=16 + 16√2, and since a=b, it is a+a+c=16 + 16√2 from the pythagorean theorem, a^2+b^2=c^2, but since it is a isosceles and a=b, we re-write this as a^2+a^2=c^2

Thanks everybody, I really appreciate it. Any luck with the first question? (For every positive even integer n, the function h(n) is defnied 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...)

Thanks everybody, I really appreciate it. Any luck with the first question? (For every positive even integer n, the function h(n) is defnied 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...)

For every positive even integer n, the function h(n) is defnied 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

Any help on how to get the correct answer? Thanks!

Answer: greater than 40

Welcome to the Gmat Club. Below is the solution for your question:

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

Answer: More than 50.

Hope it helps.

P.S. Can you please: post one question per topic, tag the questions you post, and also post the whole questions with answer choices. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...