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

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Manager
Joined: 20 Oct 2013
Posts: 66
Followers: 0

Kudos [?]: 0 [0], given: 27

Re: For every positive even integer n, the function h(n) is [#permalink]  04 May 2014, 21:27
hi

i did the same thing as karishma and when i got 2551, i thought that cannot be correct as the answer mentions above 40, so my solution must be wrong .. anyway i clicked e and it turned out right.

Bunnel & Karishma: how can u guys solve all qs so easily?
_________________

Hope to clear it this time!!
GMAT 1: 540
Preparing again

Math Expert
Joined: 02 Sep 2009
Posts: 28731
Followers: 4576

Kudos [?]: 47168 [0], given: 7108

Re: For every positive even integer n, the function h(n) is [#permalink]  22 Jun 2014, 08:39
Expert's post
Game 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 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

Merging similar topics. Please refer tot he discussion on page 1.

Check questions about various functions in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functios

Hope it helps.
_________________
BSchool Forum Moderator
Status: Happy Lunar New Year GClubbers!
Joined: 20 Apr 2014
Posts: 444
Location: Malaysia
Concentration: Strategy, Other
Schools: Ross '17 (M)
GMAT 1: 760 Q50 V41
GPA: 3.2
WE: Project Management (Other)
Followers: 10

Kudos [?]: 128 [0], given: 753

Re: For every positive even integer n, the function h(n) is [#permalink]  29 Jun 2014, 21:10
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

Q48 with 13 incorrect! you must have had a great start and screwed up towards the end, I guess...
'coz I scored a Q50 with 7 wrong answers which were evenly spread ..
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Senior Manager
Joined: 28 Apr 2014
Posts: 291
Followers: 0

Kudos [?]: 26 [0], given: 46

Re: For every positive even integer n, the function h(n) is [#permalink]  28 Jul 2014, 01:06
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.

I have the same observation. Is this the case in actual GMAT as well ? I also scored 48 on my quant inspite of a few incorrect answers. How will the real GMAT be ?

The verbal was quite difficult as well in the GMATPrep exam
BSchool Forum Moderator
Status: Happy Lunar New Year GClubbers!
Joined: 20 Apr 2014
Posts: 444
Location: Malaysia
Concentration: Strategy, Other
Schools: Ross '17 (M)
GMAT 1: 760 Q50 V41
GPA: 3.2
WE: Project Management (Other)
Followers: 10

Kudos [?]: 128 [0], given: 753

Re: For every positive even integer n, the function h(n) is [#permalink]  28 Jul 2014, 01:41
himanshujovi wrote:
I have the same observation. Is this the case in actual GMAT as well ? I also scored 48 on my quant inspite of a few incorrect answers. How will the real GMAT be ?

The verbal was quite difficult as well in the GMATPrep exam

Hi Himanshu
I found quant pretty easy. If you have appeared for CAT and scored well there (I'm talking 95 percentile+), then you'll be at ease here as well.
In case you're not comfy with verbal, then I strongly suggest that you take the free sessions from e-GMAT. They also have quite a lot free resource on GMATClub. I used their free stuff for my practice & doubt clearance and was able to score V41. To me it doesn't sound like a coincidence.

Cheers to you and all the best.
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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5735
Location: Pune, India
Followers: 1444

Kudos [?]: 7571 [0], given: 186

Re: For every positive even integer n, the function h(n) is [#permalink]  28 Jul 2014, 20:32
Expert's post
himanshujovi wrote:
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.

I have the same observation. Is this the case in actual GMAT as well ? I also scored 48 on my quant inspite of a few incorrect answers. How will the real GMAT be ?

The verbal was quite difficult as well in the GMATPrep exam

I have known people, who expect to score around 650, to get stuck on first or second question - not many but some. So be prepared for anything. Don't worry much about number of incorrect responses. GMAT will be able to judge your ability as long as you don't mislead it by wasting too much time on 1 or 2 questions or making too many careless mistakes.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5735 Location: Pune, India Followers: 1444 Kudos [?]: 7571 [1] , given: 186 Re: For every positive even integer n, the function h(n) is [#permalink] 28 Sep 2014, 23:00 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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 Responding to a pm: Quote: The question boil downs to this equation: 2^50*50! Now, let's assume this number is n. We know this number is even. So n+1 will have no factors common to n except 1. But how can we say that n+1 will have factors greater than factors of n? In this question we say that since n has prime factors from 2 to 47, n+1 will have prime factors greater than 47. This part of the solution has left me stumbled. Let me quote an example: for eg. 34 & 35 34 = 17*2 35 = 5*7 Even though 35 does not share any common factors with 34, it has prime factors less than the largest prime factor of 34 i.e. 17. So how can we be sure and say 2^2*50!+1 will have the least prime factor greater than 47? As per my understanding, since 2^50*50! has eaten up all the prime factors from 2 to 47, the only possible factors left for 2^50*50!+1 are greater than 47. Otherwise, 2^50*50!+1 it will end up having prime factors common to 2^50*50!, which is not possible. Can you help me ensure that my understanding is correct? You are comparing factors of n with factors of (n+1). You actually have to compare factors of n! with factors of n! + 1. Say n = 4. n! = 1*2*3*4 = 24 n! + 1 = 25 Will 25 have any factors common with n!? No, because n! has all factors from 1 to n. n! + 1 has 5 as a factor which is larger than n. Similarly, say n = 5 n! = 1*2*3*4*5 = 120 n! + 1 = 121 Will 121 have any factors from 1 to 5? No. All these numbers are factors of n! so they cannot be factors of n!+1. Does this make sense? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: For every positive even integer n, the function h(n) is [#permalink]  20 Oct 2014, 21:21
Shouldn't 1 be the smallest prime factor of h(100) +1?

Cheers
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5735
Location: Pune, India
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Kudos [?]: 7571 [0], given: 186

Re: For every positive even integer n, the function h(n) is [#permalink]  20 Oct 2014, 21:46
Expert's post
annie2014 wrote:
Shouldn't 1 be the smallest prime factor of h(100) +1?

Cheers

1 is not a prime number. The smallest prime number is 2.
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EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
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GMAT 1: 800 Q51 V49
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Kudos [?]: 835 [2] , given: 53

Re: For every positive even integer n, the function h(n) is [#permalink]  26 Feb 2015, 12:36
2
KUDOS
Expert's post
Hi All,

I've been asked to post this solution here, so here's another way to handle 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:

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 main idea behind this prompt is:

"The ONLY number that will divide into X and (X+1) is 1."

In other words, NONE of the factors of X will be factors of X+1, EXCEPT for the number 1.

Here are some examples:
X = 2
X+1 = 3
Factors of 2: 1 and 2
Factors of 3: 1 and 3
ONLY the number 1 is a factor of both.

X = 9
X+1 = 10
Factors of 9: 1, 3 and 9
Factors of 10: 1, 2, 5 and 10
ONLY the number 1 is a factor of both.
Etc.

Since the H(100) is (100)(98)(96)....(4)(2)....we can deduce....
1) This product will have LOTS of different factors
2) NONE of those factors will divide into H(100) + 1.

H(100) contains all of the primes from 2 through 47, inclusive (the 47 can be "found" in the "94"), so NONE of those will be in H(100) + 1. We don't even have to calculate which prime factor is smallest in H(100) + 1; we know that it MUST be a prime greater than 47....and there's only one answer that fits.

[Reveal] Spoiler:
E

GMAT assassins aren't born, they're made,
Rich
_________________

Math Expert
Joined: 02 Sep 2009
Posts: 28731
Followers: 4576

Kudos [?]: 47168 [0], given: 7108

Re: For every positive even integer n, the function h(n) is [#permalink]  26 Feb 2015, 12:38
Expert's post
EMPOWERgmatRichC wrote:
Hi All,

I've been asked to post this solution here, so here's another way to handle 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:

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 main idea behind this prompt is:

"The ONLY number that will divide into X and (X+1) is 1."

In other words, NONE of the factors of X will be factors of X+1, EXCEPT for the number 1.

Here are some examples:
X = 2
X+1 = 3
Factors of 2: 1 and 2
Factors of 3: 1 and 3
ONLY the number 1 is a factor of both.

X = 9
X+1 = 10
Factors of 9: 1, 3 and 9
Factors of 10: 1, 2, 5 and 10
ONLY the number 1 is a factor of both.
Etc.

Since the H(100) is (100)(98)(96)....(4)(2)....we can deduce....
1) This product will have LOTS of different factors
2) NONE of those factors will divide into H(100) + 1.

H(100) contains all of the primes from 2 through 47, inclusive (the 47 can be "found" in the "94"), so NONE of those will be in H(100) + 1. We don't even have to calculate which prime factor is smallest in H(100) + 1; we know that it MUST be a prime greater than 47....and there's only one answer that fits.

[Reveal] Spoiler:
E

GMAT assassins aren't born, they're made,
Rich

__________

Thank you, Rich!
_________________
Re: For every positive even integer n, the function h(n) is   [#permalink] 26 Feb 2015, 12:38

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