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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Struggling with this GMAT Prep Quant Question - any help? tagged by: Brent@GMATPrepNow ##### This topic has 4 expert replies and 0 member replies ## Struggling with this GMAT Prep Quant Question - any help? 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 ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13040 messages Followed by: 1251 members Upvotes: 5254 GMAT Score: 770 Quote: 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. Answer = E Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10197 messages Followed by: 497 members Upvotes: 2867 GMAT Score: 800 Hi adwoleonie, For future reference, you will likely receive more of a response if you post your questions directly into the appropriate sub-Forum. The PS Forum can be found here: http://www.beatthegmat.com/problem-solving-f6.html This particular Ps question shows up periodically in the Forum here. 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 except for the number 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: E GMAT assassins aren't born, they're made, Rich ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2950 messages Followed by: 19 members Upvotes: 43 adwoleonie 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 We are given that h(n) is defined to be the product of all the even integers from 2 to n, inclusive. For example, h(8) = 2 x 4 x 6 x 8. We need to determine the smallest prime factor of h(100) + 1. Before determining the smallest prime factor of h(100) + 1, we must recognize that h(100) and h(100) + 1 are consecutive integers, and consecutive integers will never share the same prime factors. Thus, h(100) and h(100) + 1 must have different prime factors. However, rather than determining all the prime factors of h(100), let’s determine the largest prime factor of h(100). Since h(100) is the product of the even integers from 2 to 100 inclusive, let’s find the largest even integer less than 100 that is the product of 2 and a prime number. That prime number is 47, since 2 x 47 = 94, which is less than 100. The next prime after 47 is 53, and 2 x 53 = 106, which is greater than 100. Therefore, 47 is the largest prime number that is a factor of h(100). In fact, all prime numbers from 2 to 47 are included in the prime factorization of h(100). Since we have mentioned that h(100) + 1 will not have any of the prime factors of h(100), all the prime factors in h(100) + 1, including the smallest one, must be greater than 47. Looking at the answer choices, only choice E can be the correct answer. Answer: E ### GMAT/MBA Expert GMAT Instructor Joined 12 Sep 2012 Posted: 2635 messages Followed by: 117 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 I like Brent's method above, but if you find yourself in a pinch on test day without a conceptual solution, you could always look for a pattern here. Suppose I take smaller versions of the function given, like h(2), h(4), and h(6). h(2) = 2 h(2) + 1 = 3 h(4) = 2 * 4 h(4) + 1= 2 * 4 + 1 = 9 h(6) = 2 * 4 * 6 h(6) + 1 = 2 * 4 * 6 + 1 = 49 So the smallest prime factor of h(2) is 3, the smallest prime factor of h(4) + 1 is 3, and the smallest prime factor of h(6) + 1 is 7. Looking for a pattern here, I notice that the smallest prime factor of h(n) + 1 seems to be either close to n or greater than n, so the answer is probably something large. With that in mind, I'd take E. I wouldn't be extremely confident, but I'd like my answer much more than I would if I randomly guessed, and I'd feel pretty good about the process. Those 70-80% confidence answers tend to add up over the course of a difficult CAT, so don't be afraid to roll up your sleeves and dig through some numbers for an answer if the conceptual approach doesn't come to you. Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. 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