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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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Struggling with this GMAT Prep Quant Question - any help?

Post Sun May 14, 2017 4:06 am
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 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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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

_________________

Scott Woodbury Stewart Founder & CEO
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5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Matt@VeritasPrep GMAT Instructor
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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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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. Find a class now!

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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

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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