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what is the smallest prime factor?

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by Jay@ManhattanReview » Thu May 04, 2017 12:59 am
Kuros 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

Source - OG diagnostic test
We have h(100) = 2.4.6.8.10....98.100 = 2^50*(1.2.3.4.5...50) = 2^50*(50!)

We see that h(100) and [h(100) + 1] are consecutive integers. Note that two consecutive integers are co-prime; it means that they do not share any common factor other than 1. For example, 50 and 51 are consecutive integers and share only 1 as a common factor.

Since h(100) = 2^50*(50!) = 2^50*(1.2.3....49.50) has all the factors from 1 to 50, it implies that [h(100) + 1] would have '1' and 'more than 50' as its factors.

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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by GMATGuruNY » Thu May 04, 2017 4:16 am
I posted an explanation here:
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by Brent@GMATPrepNow » Thu May 04, 2017 4:19 am
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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by Kuros » Thu May 04, 2017 8:19 am
Yeah I hadn't known about the co-prime rule and got stopped dead once I worked it to 2^50*(50!), thank you guys very much!
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by [email protected] » Thu May 04, 2017 8:23 am
Hi Kuros,

This question shows up every so often in this Forum and it is definitely tougher than a typical GMAT question.

As a general rule, Quant questions are almost always based on a pattern of some kind (math formula, math rule, Number Property, etc.). If you can't immediately deduce a pattern, then you might have to "play around" a bit with the question to try to deduce what the pattern is. In the broad sense, it's critical thinking: here's a weird situation - what can I do to figure it out?

Based on the description of the function in the prompt, we can run some "TESTS" to try to figure things out....

The H(n) is the product of all the even integers from 2 to n, inclusive.

So....
H(4) = 2x4 = 8
The prime factors of 8 are (2)(2)(2)
If we do H(4) + 1 = 9, then the prime factors are (3)(3)
NOTICE how NONE of the prime factors of 8 are in 9? That's interesting....

H(6) = 2x4x6 = 48
The prime factors of 48 are (2)(2)(2)(2)(3)
If we do H(6) + 1 = 49, then the prime factors are (7)(7)
NOTICE how NONE of the prime factors of 48 are in 49? That's interesting....and probably a pattern, since it's happened TWICE NOW.

From here, I'd have to deduce that this pattern holds true. With H(100), I know that there are LOTS of primes that go in (the largest of which is 47, which can be "found" in 94). I have to assume that NONE of them will go into H(100) + 1. Thus, the smallest prime would have to be greater than 47. The question doesn't actually ask us for the exact answer though (the answer choices are "ranges").

The takeaway from all of this is that you shouldn't be afraid to "play" with a question a bit. In the end, you don't have to be a brilliant mathematician to answer this question, but you're also not allowed to just stare at it either.

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by DavidG@VeritasPrep » Thu May 04, 2017 8:53 am
And if you really can't get enough of the topic, I wrote a blog post about the concept here: https://www.veritasprep.com/blog/2016/0 ... -the-gmat/
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Re: what is the smallest prime factor?

by Scott@TargetTestPrep » Mon Apr 13, 2020 3:52 am
Kuros wrote: ↑
Thu May 04, 2017 12:21 am
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a) Between 2 and 10
b) Between 10 and 20
c) Between 20 and 30
d) Between 30 and 40
e) Greater than 40

Source - OG diagnostic test
We are given that h(n) is defined as 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 prime number such that 2 times that prime number is less than 100.

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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