Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 08:20

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is the product of four consecutive integers a multiple of

Author Message
TAGS:
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1269
Followers: 23

Kudos [?]: 158 [0], given: 0

Is the product of four consecutive integers a multiple of [#permalink]  09 Jul 2006, 03:13
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Is the product of four consecutive integers a multiple of 120?

(1) None of the four digits is a multiple of 10.
(2) The sum of the four integers is a multiple of 10.
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1269
Followers: 23

Kudos [?]: 158 [0], given: 0

What kind of explanation is that?
Manager
Joined: 04 Jul 2006
Posts: 57
Followers: 1

Kudos [?]: 2 [0], given: 0

B.
interesting question

(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.

Therefore, B is the correct answer.

Last edited by game over on 10 Jul 2006, 08:41, edited 1 time in total.
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 1 [0], given: 0

I have a feeling it should be (C), but can prove this completely

Let's write this as n(n+1)(n+2)(n+3) divisible by 120

Stem1. Let's pick 1,2,3,4 as for consecutive integers. Their product is 24, not divisible by 120.

On the other hand, 2*3*4*5 is divisible by 120.

/NOT SUFF/

Stem2. Pick the same 1,2,3,4 because their sum is equal to 10. Again, NOT SUFF

I think when you combine two statements, it should be enough to prove that the product is NOT divisible by 120, but I don't want to spend my time on it right now
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 1 [0], given: 0

After reading game over's solution, I agree it should be (B).
Thanks!
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 42 [0], given: 0

Great insight!!

Took me a while to get it, but finally got it.

I was going with E as I couldn't get anywhere with the Statement 2.

Thanks/

game over wrote:
B.
interesting question

(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.

Therefore, B is the correct answer.
Similar topics Replies Last post
Similar
Topics:
2 Is the product of four consecutive even integers positive? 4 23 Jul 2010, 08:40
Is the product of four consecutive even integers positive? 3 02 Mar 2008, 08:01
What is the sum of four consecutive odd integers? 1. Product 4 13 Nov 2007, 05:51
Is the product of four consecutive even integers positive? 7 30 Jul 2007, 21:39
Is the product of six consecutive positive integers a 4 15 Jul 2006, 10:41
Display posts from previous: Sort by