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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

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

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.

Re: For every positive even integer n, the function h(n) is [#permalink]
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 .. _________________

Bored of all the GMAT and MBA stuff? Check this page out -Odds of meeting your spouse at MBA- it might give you something to laugh on or may be to be hopeful about...

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

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. _________________

Bored of all the GMAT and MBA stuff? Check this page out -Odds of meeting your spouse at MBA- it might give you something to laugh on or may be to be hopeful about...

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. _________________

Re: For every positive even integer n, the function h(n) is [#permalink]
28 Sep 2014, 23:00

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Expert's post

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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.

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