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

It is currently 27 Aug 2014, 13:19

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If n is a positive integer and the product of all the

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1128
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 23 [0], given: 0

GMAT Tests User
If n is a positive integer and the product of all the [#permalink] New post 30 Jan 2005, 17:29
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

A. 8
B. 9
C. 10
D. 11
E. 12

Please explain the way to find it.
Thanks.
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

GMAT Tests User
 [#permalink] New post 30 Jan 2005, 18:42
"D".

convert 990 in prime factors = 11x3x3x5x2

u will see that for the product to be divisble by 990....product shud atleast contain a 11.....so n shud go upto 11 .....so least n = 11
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 196 [0], given: 0

GMAT Tests User
 [#permalink] New post 30 Jan 2005, 19:22
What if the number is 189 instead of 990?
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

GMAT Tests User
 [#permalink] New post 30 Jan 2005, 19:30
HongHu wrote:
What if the number is 189 instead of 990?


good one buddy....shud be "9".
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 196 [0], given: 0

GMAT Tests User
 [#permalink] New post 30 Jan 2005, 20:17
You got it. :-D
Intern
Intern
avatar
Joined: 31 Jan 2005
Posts: 4
Followers: 0

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

 [#permalink] New post 31 Jan 2005, 08:06
banerjeea_98 wrote:
HongHu wrote:
What if the number is 189 instead of 990?


good one buddy....shud be "9".


I just have a quick question, why is it 9 and not 7?

I was wondring because 189=7*3*3*3 so I thought the answer was going to be 7. :?
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 196 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 08:32
From 1 to seven you've only got two 3s, one from 3, another from 6, but you need three 3s. :-D That's why I used this number 189, to set the trap. ;)
Director
Director
User avatar
Joined: 21 Sep 2004
Posts: 619
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 09:07
HongHu wrote:
From 1 to seven you've only got two 3s, one from 3, another from 6, but you need three 3s. :-D That's why I used this number 189, to set the trap. ;)

Honghu, can you please explain in detail how u solved the question. thanks.
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3405
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 158 [0], given: 2

GMAT Tests User
Re: product and integer n [#permalink] New post 31 Jan 2005, 10:32
I looked at it, and broke 990 to its prime, then 11*5*3*3*2

i notice that the only prime number that shows up is 11, therefor it will be the smallest from this selection, but I think techically the smallest possible value should be 2?

Honghu can you please explain your reasoning with 189?



Antmavel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

A. 8
B. 9
C. 10
D. 11
E. 12

Please explain the way to find it.
Thanks.
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1498
Followers: 6

Kudos [?]: 27 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 11:06
another good question to be discussed and put my 2 cts views.

n to be divided by 990, n must posses all the prime factors of 990. the prime factors of 990 are 2, 3, 3, 5, and 11. the question stipulated that n is a product of all positive integers 1 to n means no single integer can repete if we need one more integer in between 1 to n. for example, here we need one more 3, we have to include 6 to fulfill the requirement of 3. similarly, to fulfill the requirement to 11, we have to stretch the n up to 11.

in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.
Director
Director
User avatar
Joined: 21 Sep 2004
Posts: 619
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 12:13
thank you MA. thats a very good explanation indeed.
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3405
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 158 [0], given: 2

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 12:22
can you explain what you understood? I still have a hard time following...

vprabhala wrote:
thank you MA. thats a very good explanation indeed.
Intern
Intern
avatar
Joined: 31 Jan 2005
Posts: 4
Followers: 0

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

 [#permalink] New post 31 Jan 2005, 12:25
MA wrote:
another good question to be discussed and put my 2 cts views.



in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.


Oh!!! Now I get it! Thanks for the explination!
Director
Director
avatar
Joined: 27 Dec 2004
Posts: 910
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 13:48
I don't get this folks please help me :oops: :cry:
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3405
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 158 [0], given: 2

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 18:03
guys explain this to me please!

I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 18:43
fresinha12 wrote:
guys explain this to me please!

I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?


ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ?
However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.

product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? :-D [/b]
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3405
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 158 [0], given: 2

GMAT Tests User
 [#permalink] New post 31 Jan 2005, 19:11
got it....thanks!

banarjee nice explanation...

banerjeea_98 wrote:
fresinha12 wrote:
guys explain this to me please!

I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?


ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ?
However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.

product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? :-D [/b]
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1128
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 23 [0], given: 0

GMAT Tests User
 [#permalink] New post 01 Feb 2005, 20:25
Honnestly the first explanation from MA was really great and banerjeea_98 finished the work in beauty :-D Thanks guys, I totally understand now what I totally didn't understand 5mn ago....

Princeton and Kaplan are useless compare to your explanations
:lol:
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1498
Followers: 6

Kudos [?]: 27 [0], given: 0

GMAT Tests User
 [#permalink] New post 01 Feb 2005, 23:05
Thanks guys. i am truly overwhelmed.........
Manager
Manager
avatar
Joined: 24 Jan 2005
Posts: 217
Location: Boston
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
 [#permalink] New post 04 Feb 2005, 11:51
D as only way to have 990 is to have 11 which is prime
  [#permalink] 04 Feb 2005, 11:51
    Similar topics Author Replies Last post
Similar
Topics:
If n is a positive integer and the product of all the tarek99 3 31 Mar 2008, 05:43
Experts publish their posts in the topic If n is a positive integer and the product of all the bsjames2 5 05 Jan 2008, 13:30
If N is a positive integer, and the product of all the eazyb81 3 17 Sep 2006, 12:57
If n is a positive integer and the product of all the yach 3 06 Mar 2006, 15:11
If n is a positive integer and the product of all the TOUGH GUY 1 10 Dec 2005, 15:16
Display posts from previous: Sort by

If n is a positive integer and the product of all the

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.