Last visit was: 24 Apr 2026, 07:53 It is currently 24 Apr 2026, 07:53
Home
Messages
Tests
Schools
Events
Close
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
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
lonesome307
User avatar
Joined: 12 Jun 2006
Last visit: 14 Dec 2007
Posts: 44
Own Kudos:
12
Posts: 44
Kudos: 12
Gmat Prep PS 19 Sep 2007, 22:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

History

Date
Time
Result
Not Attempted Yet
Hi All,

Can someone please help me solve the following question?

Thanks



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
GMATBLACKBELT
User avatar
Joined: 29 Mar 2007
Last visit: 03 Jun 2013
Posts: 1,138
Own Kudos:
1,913
Posts: 1,138
Kudos: 1,913
Gmat Prep PS 19 Sep 2007, 22:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Im really confused with this question as well. Ok so h(n) is the product of all the even integers from 2-n.

n=100 so all the evens from 2-100. 100-2+1=99. Now 98/2=49+1 = 50 evens from 100-2.

so these (50 evens* eachother) + 1. This is going to be an odd number...

if p is the smallest prime factor... I don't know how to find the product of all the even numbers in a situation like this.

I can find sum... how do u find the product? And then from there, how do u find out whether the prime factor is greater than 40????
User avatar
Ferihere
User avatar
Joined: 27 Aug 2007
Last visit: 03 Sep 2018
Posts: 121
Own Kudos:
28
Posts: 121
Kudos: 28
Gmat Prep PS 20 Sep 2007, 10:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1

so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)
User avatar
ashkrs
User avatar
Joined: 08 Jun 2007
Last visit: 21 Feb 2019
Posts: 282
Own Kudos:
345
GMAT 1: 680 Q48 V35
Posts: 282
Kudos: 345
Gmat Prep PS 21 Sep 2007, 10:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)
Show more


Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks
User avatar
Ferihere
User avatar
Joined: 27 Aug 2007
Last visit: 03 Sep 2018
Posts: 121
Own Kudos:
28
Posts: 121
Kudos: 28
Gmat Prep PS 22 Sep 2007, 07:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ashkrs
Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)

Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks
Show more


There are 2 to 100 even numbers, so that means all of them divisible by 2. You can rewrite them as (2*1)*(2*2)*(2*3)*(2*4)*....*(2*50) + 1
as you can see there are 50 2's, that is why 2^50 (I wrote it down as 2^100, but still it will not change the asnwer) and other numbers just 1*2*3*4*....*50 so we can rewrite it 2^50*(1*2*3*4*....*50)+1

Is it clear?
User avatar
GMATBLACKBELT
User avatar
Joined: 29 Mar 2007
Last visit: 03 Jun 2013
Posts: 1,138
Own Kudos:
1,913
Posts: 1,138
Kudos: 1,913
Gmat Prep PS 24 Sep 2007, 10:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ferihere
ashkrs
Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)

Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks

There are 2 to 100 even numbers, so that means all of them divisible by 2. You can rewrite them as (2*1)*(2*2)*(2*3)*(2*4)*....*(2*50) + 1
as you can see there are 50 2's, that is why 2^50 (I wrote it down as 2^100, but still it will not change the asnwer) and other numbers just 1*2*3*4*....*50 so we can rewrite it 2^50*(1*2*3*4*....*50)+1

Is it clear?
Show more


Very nice solution, I didn't understand this approach til just now :P



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderator:
Bunuel
Math Expert
109814 posts