GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 17:02

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# For every positive even integer n, the function h(n) is defined to be

Author Message
TAGS:

### Hide Tags

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15294
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

02 Jan 2015, 13:10
6
Hi,

This question tends to trouble most Test Takers 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 prime number though - it just asked for the range that the prime would fall into.

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.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15294
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

26 Feb 2015, 13:36
10
Hi All,

I've been asked to post this solution here, so here's another way to handle 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:

1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40

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.

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.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

12 Nov 2015, 09:06
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

As per the definition of the question

h(100) = 2 x 4 x 6 x 8 x 10 x 12 x 14 ... and so on...98 x 100 (Total 50 terms)

=> h(100) = 2^50 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10......and so on...x 48 x 49 x 50)

This means h(100) is multiple of all prime numbers between 1 and 50

therefore h(100)+1 will leave a remainder of 1 when divided by any prime number from 1 to 50

therefore, p, which is a factor of h(100)+1, will certainly be greater than a prime numbers greater than 50

Hence, "p" must be greater than 40 as per the following options

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Math Expert
Joined: 02 Aug 2009
Posts: 7991
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

31 Jan 2016, 01:07
2
gnikhilreddy143 wrote:
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, 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

Hi,
without any calculations we can make out that h(n) will consist of products of all numbers till 50and 2^50, as it is even integers so 100/2..
therefore when we add 1 to this product , it will not be a multiple of any number till 50..
so the smallest prime factor has to be >50..
E..

lets solve it..
h(100)= 2*4*6...*100..
= 2*1*2*2*2*3...2*50= 2^50*1*2*...50..
h(100)+1= 2^50*1*2*...50 +1..
so prime factor>50..
ans E >40

NOTE:- Merging topic. Please type the first few letters as TOPIC name correctly
_________________
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4015
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

31 Jan 2016, 01:30
4
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.

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4015
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

26 Nov 2016, 13:43
Top Contributor
blazov 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, than 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

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.

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
GMAT Tutor
Joined: 01 Oct 2016
Posts: 9
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

05 Jan 2017, 22:39
1
H(100) will be a multiple of all of the prime factors below 50 because it is the product of all of the primes below 50 multiplied by 2 (94 is 47 x 2, 62 is 31 x 2, etc. If h(100) is a multiple of each of these, then when we add 1 it will throw us off of being a multiple of all of these, so the greatest prime factor will have to be greater than 50, which is greater than 40.
_________________
Dan the GMAT Man
Offering tutoring and admissions consulting in the NYC area and online
danthegmatman.squarespace.com
danthegmatman@gmail.com
Senior Manager
Joined: 17 Mar 2014
Posts: 429
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

21 Apr 2017, 11:22
One of the best question and best explanation by Bunuel.

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 co-prime, 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.

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1348
Location: Malaysia
For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

01 May 2017, 05:36
Bunuel 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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, 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.

Dear Bunuel, if the question changed to positive ODD integer, will the answer be the same?

I think answer choice A should be between 2 and 10.
_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Math Expert
Joined: 02 Sep 2009
Posts: 58434
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

01 May 2017, 05:49
ziyuen wrote:
Bunuel 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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, 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.

Dear Bunuel, if the question changed to positive ODD integer, will the answer be the same?

I think answer choice A should be between 2 and 10.

Yes. 49th odd number is 97, which is prime.
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8117
Location: United States (CA)
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

20 Jun 2017, 07:37
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

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 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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Joined: 21 Jan 2017
Posts: 31
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

26 Jul 2017, 02:51
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 co-prime, 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.

Hi Bunuel,

Thanks for the explanation! It was helpful.
I tried to apply the same logic with simple numbers. Consider we have 15. So if try to find out the least prime factor of 15 , we must take 14 and check for its prime factors. They are 2 and 7. So according to the above theory, will the least prime factor of 15 be > 7??? It should not! ( here, do we need to check >2 or >7 ??)

Where am I going wrong??

Thanks,
Uma
Math Expert
Joined: 02 Sep 2009
Posts: 58434
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

26 Jul 2017, 04:24
umabharatigudipalli wrote:
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 co-prime, 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.

Hi Bunuel,

Thanks for the explanation! It was helpful.
I tried to apply the same logic with simple numbers. Consider we have 15. So if try to find out the least prime factor of 15 , we must take 14 and check for its prime factors. They are 2 and 7. So according to the above theory, will the least prime factor of 15 be > 7??? It should not! ( here, do we need to check >2 or >7 ??)

Where am I going wrong??

Thanks,
Uma

Two consecutive numbers are co-prime, so 14 and 15 are also co-prime, they do not share any common factor but 1. The factors of 14 are 1, 2, 7, and 14 and the factors of 15 are 1, 3, 5, and 15. As you can see no common factors but 1 (including no common primes).

The reasons you are confused is that $$h(100)=2^{50}*50!$$ has ALL prime numbers from 1 to 50 as its factors, thus $$h(100)+1=2^{50}*50!+1$$ won't have ANY prime factor from 1 to 50.

But 14 does NOT have ALL primes from 1 to 14, so 15 might have some primes from 1 to 14 (but 14 won't have the same primes as 15).
_________________
Intern
Joined: 05 Oct 2017
Posts: 45
Concentration: Accounting, Social Entrepreneurship
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

29 Oct 2018, 13:05
1
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 + (n-1) 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 + (n-1) 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

Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100

By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

Hope that helps

Posted from my mobile device
_________________
Director
Joined: 24 Oct 2016
Posts: 536
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

01 Oct 2019, 17:28
Bunuel 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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, 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.

Bunuel Thanks for the solution. There seems to be a small typo in "$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$"

It should be 2*50! and not 2^50*50!
_________________

If you found my post useful, KUDOS are much appreciated. Giving Kudos is a great way to thank and motivate contributors, without costing you anything.
Math Expert
Joined: 02 Sep 2009
Posts: 58434
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

01 Oct 2019, 20:34
dabaobao wrote:
Bunuel 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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, 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.

Bunuel Thanks for the solution. There seems to be a small typo in "$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$"

It should be 2*50! and not 2^50*50!

No typo there: https://gmatclub.com/forum/for-every-po ... l#p1035831
_________________
Director
Joined: 24 Oct 2016
Posts: 536
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: For every positive even integer n, the function h(n) is defined to be  [#permalink]

### Show Tags

02 Oct 2019, 02:58
Bunuel wrote:
dabaobao wrote:
Bunuel 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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, 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.

Bunuel Thanks for the solution. There seems to be a small typo in "$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$"

It should be 2*50! and not 2^50*50!

No typo there: https://gmatclub.com/forum/for-every-po ... l#p1035831

My bad! You're right! I was looking at * as +. Sorry for the trouble. Thanks again!
_________________

If you found my post useful, KUDOS are much appreciated. Giving Kudos is a great way to thank and motivate contributors, without costing you anything.
Re: For every positive even integer n, the function h(n) is defined to be   [#permalink] 02 Oct 2019, 02:58

Go to page   Previous    1   2   [ 37 posts ]

Display posts from previous: Sort by