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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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04 May 2014, 22:27

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 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 defined to be [#permalink]

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29 Jun 2014, 22:10

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 ..
_________________

I'm happy to help if you wanna know about Ross & UMich, but please do not come to me with your GMAT issues or questions. And please add a bit of humor to your questions or you'll bore me to death.

Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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28 Jul 2014, 02:06

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 defined to be [#permalink]

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28 Jul 2014, 02:41

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.

I'm happy to help if you wanna know about Ross & UMich, but please do not come to me with your GMAT issues or questions. And please add a bit of humor to your questions or you'll bore me to death.

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.
_________________

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.

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.

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.

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

As per the definition of the question

h(100) = 2 x 4 x 6 x 8 x 10 x 12 x 14 ... and so on...98 x 100 (Total 50 terms)

=> h(100) = 2^50 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10......and so on...x 48 x 49 x 50)

This means h(100) is multiple of all prime numbers between 1 and 50

therefore h(100)+1 will leave a remainder of 1 when divided by any prime number from 1 to 50

therefore, p, which is a factor of h(100)+1, will certainly be greater than a prime numbers greater than 50

Hence, "p" must be greater than 40 as per the following options

Answer: Option E
_________________

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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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23 Dec 2015, 16:13

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.

For every positive even integer n, the function h(n) is defined to be the product of all 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

Hi, without any calculations we can make out that h(n) will consist of products of all numbers till 50and 2^50, as it is even integers so 100/2.. therefore when we add 1 to this product , it will not be a multiple of any number till 50.. so the smallest prime factor has to be >50.. E..

lets solve it.. h(100)= 2*4*6...*100.. = 2*1*2*2*2*3...2*50= 2^50*1*2*...50.. h(100)+1= 2^50*1*2*...50 +1.. so prime factor>50.. ans E >40

NOTE:- Merging topic. Please type the first few letters as TOPIC name correctly _________________

For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is

A: Between 2 & 10 B: Between 10 & 20 C: Between 20 & 30 D: Between 30 & 40 E: Greater than 40

Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1 For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1) Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let’s examine h(100) h(100) = (2)(4)(6)(8)….(96)(98)(100) = (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50) Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

. . . . Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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31 Jan 2016, 05:56

neo656 wrote:

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 ..

I have got 49 with 13 incorrect in GMAT Prep Practice Test-1. However, my verbal score was screwed up, and it was 23. I have just started my preparation.
_________________

Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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09 Nov 2016, 02:28

Hello

I didn't understand how the formula works Product of terms = Average * number of terms as per the formula h(100) = 2550

But if I multiply 2 *100 = 200 *4 = 800 * 6 = 4800 > 2550 ( I have multiplied only 4 numbers, I have to multiply 8,10...,98. SO the h(100) will be much greater ?

Am I missing anything?

Alok

VeritasPrepKarishma wrote:

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

the first part has all numbers (including prime) as it's factors. First part+1 is a consecutive number so they do not share any common factors (other than 1, which neither prime nor composite). Therefore the smallest prime factor for h(100)+1 is greater than 50.

Answer : E
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