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Struggling with this GMAT Prep Quant Question - any help?

This topic has 4 expert replies and 0 member replies
adwoleonie Newbie | Next Rank: 10 Posts Default Avatar
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11 May 2015
Posted:
1 messages

Struggling with this GMAT Prep Quant Question - any help?

Post Sun May 14, 2017 4:06 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

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    Post Sun May 14, 2017 4:39 am
    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

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    Post Sun May 14, 2017 10:56 am
    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

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    Post Thu Jun 01, 2017 4:06 pm
    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

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    Post Wed Jun 21, 2017 4:44 pm
    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.

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