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

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Re: For every positive even integer n, the function h(n) is [#permalink] New post 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?
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For every positive even integer n, the function h(n) is defi [#permalink] New post 22 Jun 2014, 08:35
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
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 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.
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 29 Jun 2014, 21:10
imadkho wrote:
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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Re: For every positive even integer n, the function h(n) is [#permalink] New post 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
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 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. :-D
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 28 Jul 2014, 20:32
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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.
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 28 Sep 2014, 23:00
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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?
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Re: For every positive even integer n, the function h(n) is [#permalink] New post 20 Oct 2014, 21:21
Shouldn't 1 be the smallest prime factor of h(100) +1?

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Re: For every positive even integer n, the function h(n) is [#permalink] New post 20 Oct 2014, 21:46
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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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Re: For every positive even integer n, the function h(n) is   [#permalink] 20 Oct 2014, 21:46
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