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For every positive even integer n, the function h(n) is defined to be [#permalink]
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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 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 + (n1) 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 + (n1) 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
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Last edited by Bunuel on 18 Oct 2016, 03:25, edited 4 times in total.
Edited the question and added the OA



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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
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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 coprime, 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. Answer: 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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Bunuel  how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain.
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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 defined to be [#permalink]
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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: For every positive even integer n, the function h(n) is defined to be [#permalink]
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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 coprime, 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. Answer: E. That is an impressive solution. < understatement



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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
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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 + (n1) 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 + (n1) 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 ... hpartii/
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
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Stiv, Here are some other official GMAT questions that are conceptually related: 1) GMAT Prep: doestheintegerkhaveafactorpsuchthat1pk126735.html2) Official Guide GMAT 13th Edition: http://www.beatthegmat.com/divisibilityt111432.htmlVideo explanation: http://www.gmatquantum.com/og13/77prob ... ition.html3) Video explanation to the original problem(For every positive even integer n, the function h(n) is....): http://www.gmatquantum.com/sharedposts ... lem13.htmlDabral
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
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Stiv wrote:
Karishma, could you recommend me some other online material or sites where I can study these kind of problems indepth? 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: ifnisapositiveintegerandristheremainderwhen119518.html?hilit=factor%20consecutive%20integers
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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. this was my second question on the first practice test, i'll have to review this answer quite a few times until it sticks. Great answer



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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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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.
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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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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 coprime 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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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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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 ... hpartii/
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
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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 (p1) when it divides h(100). remainder(h(100)/p) = p1. 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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Responding to a pm: Quote: The question here talks about h(50)+1
Last term number n=25 Average = 26 Sum = 26*25=650
h(50)+1 = 651 which is not a prime number as it is divisible by 3. Please correct me if i m wrong here. my answer is A.
I am not sure where you got this question since the original question here talks about h(100) + 1. Even if we do assume h(50) + 1, note that h(2) = 2 h(4) = 2*4 h(6) = 2*4*6 .... So I am not sure how you got average 26. The correct answer here is (E) as explained in the link given in my post above.
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