GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 03 Apr 2020, 15:44

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Which of these must the factor of the product of four consecutive even

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Current Student
User avatar
D
Joined: 12 Aug 2015
Posts: 2537
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User
Which of these must the factor of the product of four consecutive even  [#permalink]

Show Tags

New post 05 Aug 2016, 20:02
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:59) correct 44% (02:02) wrong based on 39 sessions

HideShow timer Statistics

Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them

_________________
Manager
Manager
avatar
Joined: 21 Jun 2016
Posts: 69
Location: India
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

Show Tags

New post 05 Aug 2016, 22:20
Answer is D

80 is not factor.... It has to have one 5.

Sent from my Lenovo A7000-a using GMAT Club Forum mobile app
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 11 Sep 2015
Posts: 4579
Location: Canada
GMAT 1: 770 Q49 V46
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

Show Tags

New post 08 Feb 2020, 07:53
Top Contributor
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them


Let \(2k =\) the smallest EVEN integer [all even numbers can be written as the product 2k, where K is an integer]
So, \(2k +2 =\) the next EVEN integer
\(2k +4 =\) the next EVEN integer
And \(2k +6 =\) the greatest EVEN integer

So the product of these integers \(= (2k)(2k+2)(2k+4)(2k+6)\)
\(= (2k)[2(k+1)][2(k+2)][2(k+3)]\)
\(= (2)(2)(2)(2)(k)(k+1)(k+2)(k+3)\)

Important: notice that k, k+1, k+2 and k+3 are FOUR consecutive integers

Important property: There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

Since k, k+1, k+2 and k+3 are FOUR consecutive integers, we know that one of those values is divisible by 2, and another value is divisible by 4
In other words, we can factor a 2 out of one of the integers, and we can factor of 4 out of one of the integers.
We can also factor a 3 out of one of the integers.

So our product becomes: \((2)(2)(2)(2)(2)(4)(3)(?)(?)(?)\)

Now let's check the five answer choices...
1) 48 = (2)(2)(2)(2)(3), since there are four 2's and one 3 "hiding" in our product, we know that the product must be divisible by 48
2) 64 = (2)(2)(2)(2)(2)(2), since there are six 2's "hiding" in our product, we know that the product must be divisible by 64
3) 96 = (2)(2)(2)(2)(2)(3), since there are fix 2's and one 3 "hiding" in our product, we know that the product must be divisible by 96
4) 192 = (2)(2)(2)(2)(2)(2)(3), since there are six 2's and one 3 "hiding" in our product, we know that the product must be divisible by 192
5) 80 = (2)(2)(2)(2)(5), since we can't be certain that there's a 5 "hiding" in our product, we can't be certain that the product is divisible by 80

Answer: D

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Image
CEO
CEO
User avatar
V
Joined: 03 Jun 2019
Posts: 2502
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Premium Member Reviews Badge CAT Tests
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

Show Tags

New post 08 Feb 2020, 08:09
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them


Product of four consecutive integers will be of the form = 2x*2(x+1)*2(x+2)*2(x+3) = 16*x(x+1)(x+2)(x+3); where x is an integer
x(x+1)(x+2)(x+3) will be divisible by 4*2*3 = 24
16*x(x+1)(x+2)(x+3) will be divisible by 24*16= 2^7*3

1) 48 = 2^4*3 is a factor of 2^7*3
2) 64 = 2^6 is a factor of 2^7*3
3) 96 = 2^5*3 is a factor of 2^7*3
4) 192 = 2^6*3 is a factor of 2^7*3
5) 80 = 2^5*5 is NOT a factor of 2^7*3

IMO D
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1242
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

Show Tags

New post 08 Feb 2020, 09:32
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them


product of least such sequence=2*4*6*8=384
384 divisible by 48, 64, 96, 192
D
GMAT Club Bot
Re: Which of these must the factor of the product of four consecutive even   [#permalink] 08 Feb 2020, 09:32
Display posts from previous: Sort by

Which of these must the factor of the product of four consecutive even

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne