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

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For every positive even integer n, the function h(n) is [#permalink]  28 Jan 2012, 16:46
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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

[Reveal] Spoiler:
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.

h(100) = 2 * 4 * 6 ****************100

Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.

100 = 2 + (n-1) 2
n = 50

Product of terms = Average * number of terms

Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.

H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
[Reveal] Spoiler: OA

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Last edited by Bunuel on 06 Jul 2013, 01:27, edited 3 times in total.
Edited the question and added the OA
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Re: Range of p and least prime factor [#permalink]  28 Jan 2012, 16:53
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enigma123 wrote:
h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40
< 30
> 40
< 10
Indeterminate

Below is the proper version of 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?
A. between 2 and 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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

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Re: For every positive even integer n, the function h(n) is [#permalink]  10 May 2012, 06:08
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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

This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.

h(100) = 2 * 4 * 6 ****************100

Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.

100 = 2 + (n-1) 2
n = 50

Product of terms = Average * number of terms

Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.

H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10

Check out this post for detailed theory involved in this question: http://www.veritasprep.com/blog/2011/09 ... h-part-ii/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 29090 Followers: 4720 Kudos [?]: 49594 [15] , given: 7395 Re: For every positive even integer n, the function h(n) is [#permalink] 28 Jan 2012, 17:08 15 This post received KUDOS Expert's post 6 This post was BOOKMARKED enigma123 wrote: Bunuel - how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain. Given that the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive, so: $$h(100)=2*4*6*...*100=(2*1)*(2*2)*(2*3)*(2*4)*...*(2*50)$$ --> factor out all 50 2's: $$h(100)=2^{50}*(1*2*3*..*50)=2^{50}*50!$$. Hope it's clear. _________________ Senior Manager Joined: 19 Apr 2009 Posts: 435 Location: San Francisco, California Followers: 77 Kudos [?]: 284 [8] , given: 5 Re: For every positive even integer n, the function h(n) is [#permalink] 12 May 2012, 01:45 8 This post received KUDOS 4 This post was BOOKMARKED Stiv, Here are some other official GMAT questions that are conceptually related: 1) GMAT Prep: does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html 2) Official Guide GMAT 13th Edition: http://www.beatthegmat.com/divisibility-t111432.html Video explanation: http://www.gmatquantum.com/og13/77-prob ... ition.html 3) Video explanation to the original problem(For every positive even integer n, the function h(n) is....): http://www.gmatquantum.com/shared-posts ... lem13.html Dabral Attachments image12.png [ 33.64 KiB | Viewed 60339 times ] _________________ New!2016 OFFICIAL GUIDE FOR GMAT REVIEW: Free Video Explanations. http://www.gmatquantum.com Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5850 Location: Pune, India Followers: 1478 Kudos [?]: 7925 [3] , given: 190 Re: For every positive even integer n, the function h(n) is [#permalink] 27 Aug 2012, 21:52 3 This post received KUDOS Expert's post mehulsayani wrote: Hey Bunuel, I used the following logic, and got E. (2*4*6*8.... 100) +1 is of the format 6p +1. Therefore, its a prime number and has only two factors, one and itself. So, its definitely greater than 40. So, E Is it correct? Every prime number greater than 3 is of one of the two forms: (6a -1), (6a + 1) e.g. 5 = 6*1 - 1; 7 = 6*1 + 1; 11 = 6*2 - 1; 13 = 6*2 + 1 etc But every number of the form (6a -1) or (6a + 1) is not prime. e.g. 25 = 6*4 + 1 i.e. it is of the form 6a+1. But 25 is not prime. So just because a number is of the form 6a + 1, we cannot say that it is definitely prime. _________________ 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]  21 Sep 2012, 00:49
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Got this one on the 9th question of my GMATPrep test. I actually reached a good point on this problem.

I got to the point that the product of 2*4*6*....*100 = (2^50)*50!

At this point I lost the plot and developed a weird approach here.

Since the last number in the given h(n) was 100 so the actual value could be like xxx00. So now, h(n) + 1 would be some value like xxx01.

So the possible prime factors are effectively 3, 7, 11, 13 etc.

I simply gambled on one of these till 20 being a factor and chose A. It was wrong though. The correct answer is indeed E.

The explanations above specially the one by Bunuel where the point of co-prime numbers not having any common factors but 1 was particularly good stuff and I had not read it before. This information will surely help the next time provided I am able to fully recognize it down.
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Re: GMAT Prep Question - Need help [#permalink]  22 Apr 2013, 21:45
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itsmeabhi99 wrote:
For every positive 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

I have discussed this question and the related theory in detail here: http://www.veritasprep.com/blog/2011/09 ... h-part-ii/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 3427 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 147 Kudos [?]: 922 [2] , given: 56 Re: For every positive even integer n, the function h(n) is [#permalink] 26 Feb 2015, 12:36 2 This post received 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. Final Answer: [Reveal] Spoiler: E GMAT assassins aren't born, they're made, Rich _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5850 Location: Pune, India Followers: 1478 Kudos [?]: 7925 [1] , given: 190 Re: For every positive even integer n, the function h(n) is [#permalink] 12 May 2012, 02:10 1 This post received KUDOS Expert's post Stiv wrote: Karishma, could you recommend me some other on-line material or sites where I can study these kind of problems in-depth? Thank you! Check out part i of the link given above as well. That should cover the theory that is useful for such questions. As for examples, you can search for 'factors consecutive integers' in the Quant forum and you should hit quite a few questions based on these concepts. I got one in one of my posts: if-n-is-a-positive-integer-and-r-is-the-remainder-when-119518.html?hilit=factor%20consecutive%20integers _________________ 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]  27 Aug 2012, 04:07
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Hey Bunuel, I used the following logic, and got E.
(2*4*6*8.... 100) +1 is of the format 6p +1. Therefore, its a prime number and has only two factors, one and itself. So, its definitely greater than 40.

So, E

Is it correct?
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Re: For every positive integer n, the function h(n) is defined [#permalink]  02 Sep 2012, 10:31
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I'm not taking credit for the following solution. Found it in another forum.

Two consecutive integers cannot be divisible by the same integer greater than 1.

So if we can prove that h(100) is divisible by every number smaller than 50, we proved that h(100)+1 is NOT divisible by any number smaller than 50 (besides 1).

2*4*6*...*98*100=2*(1*2*3*...*50)=2*(50!)
Hence, h(100) is divisible by every number smaller than 51. So the smallest prime factor of h(100)+1 is at least 53.
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Re: GMAT Prep Question - Need help [#permalink]  22 Apr 2013, 21:40
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h(n) = the product of all even integers from 2 to n inclusive
=> h(100) + 1 = (2 * 4 * 6 * 8 * ....... * 100) + 1
= 2^50 * (1*2*3*4*......*50) + 1

As all the numbers from 2 to 50 are factors of 2^50 * (1*2*3.....*50), none of these can be factors of 2^50 * (1*2*3*4*......*50) + 1 (i.e. of h(100)+1)

Therefore the smallest prime factor of h(100) + 1 is greater than 50.

This corresponds to option (E).
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Re: For every positive even integer n, the function h(n) is [#permalink]  12 May 2013, 11:19
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This is a great question. Let me try to post a simple solution!

Let's divide the solution into two parts:

Part (1):

We need to simplify h(100) first.

h(100) = 2*4*6*8*....*98*100.

Consider that 2 = 2*1, 4 = 2*2, etc. up to 100 = 2*50.

h(100) then simplifies to $$h(100) = 2*1 * 2*2 * 2*3 *.... * 2*49 * 2*50 = 2^{50} * 1*2*3*4*... *50$$ (A)

Part (2):

Consider the second part of the question. We need to find a bound for the smallest prime factor of h(100)+1. Any number p that is a factor of h(100)+1, leaves a remainder of (p-1) when it divides h(100).

remainder(h(100)/p) = p-1.

Another way of saying this is "p is a prime number that is not a divisor of h(100)" (B)

From (A):

Clearly h(100) is divisible by all prime numbers less than 50 (as h(100) is divisible by 50!). The smallest prime that is not a divisor of h(100) should therefore be greater than 50.

Looking at the choices, the most appropriate choice is:

(e) greater than 40.

Hope that Helps!
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Re: For every positive even integer n, the function h(n) is [#permalink]  12 Jun 2013, 04:25
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Re: For every positive even integer n, the function h(n) is [#permalink]  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]  28 Jan 2012, 17:00
Bunuel - how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain.
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Re: For every positive even integer n, the function h(n) is [#permalink]  02 Mar 2012, 12:26
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
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Re: For every positive even integer n, the function h(n) is [#permalink]  01 May 2012, 05:42
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.
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Re: Range of p and least prime factor [#permalink]  10 May 2012, 05:53
Bunuel wrote:
enigma123 wrote:
h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40
< 30
> 40
< 10
Indeterminate

Below is the proper version of 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?
A. between 2 and 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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

That is an impressive solution. <-- understatement
Re: Range of p and least prime factor   [#permalink] 10 May 2012, 05:53

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