It is currently Thu Sep 02, 2010 11:42 am
Image

Win a Prep Course, or an iPad, or a Business School Visit of Your Dreams!
Fly in First/Business class from virtually anywhere to wherever on this planet, stay in a Luxury hotel, and take a limo to the airport!

GMAT Club is at a historic milestone of 100,000 members! We are celebrating by giving away 375 prizes worth over $48,000.




Post new topic Reply to topic  [ 14 posts ]  Bookmark and Share
Author Message
TAGS:
  Remainders [#permalink]
PostPosted: Fri Oct 23, 2009 3:04 pm 
Offline
Senior Manager
Senior Manager

Joined: Thu Nov 03, 2005
Posts: 424
Location: Chicago, IL
Followers: 0

Kudos (?): 7 (0), given: 16

GMAT Tests User
         00:00        
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18

_________________
Hard work is the main determinant of success


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Fri Oct 23, 2009 11:35 pm 
Offline
Manager
Manager

Joined: Sun Jul 05, 2009
Posts: 221
Followers: 0

Kudos (?): 2 (0), given: 5

GMAT Tests User
Opps! Really a hard one. I cannot figure out how to quickly solve it!! Please help someone.


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 12:22 am 
Offline
GMAT Quant Forum Moderator
User avatar

Joined: Wed Sep 02, 2009
Posts: 2270
Followers: 91

Kudos (?): 1133 (1), given: 78

GMAT Tests User
Promotion
This one really needs some calculations and I don't know how it can be done more quickly:

Let's find the closest multiple of 33 to these numbers: it's 1056.

(1056-12)(1056-9)(1056-6)(1056-3) every term after simplification will have 1056 as its multiple except the last one which will be 12*9*6*3.

So the remainder will be the same when 12*9*6*3 is divided by 33.

The same way here 12*6*3=216 remainder when divided by 33 =18,
9*18=162 remainder=30

Done

Answer: 30

I believe there is a easier solution...

_________________
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 7:35 am 
Offline
Manager
Manager

Joined: Mon Aug 11, 2008
Posts: 204
Followers: 0

Kudos (?): 27 (0), given: 8

GMAT Tests User
I think it's easier: because all of the term divisible by 3 so the remainder of the multiply of them will be the remainder when after dividing by 3 then dividing by 11. This can only be 3 because all other choice is greater than 11


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 9:22 am 
Offline
GMAT Quant Forum Moderator
User avatar

Joined: Wed Sep 02, 2009
Posts: 2270
Followers: 91

Kudos (?): 1133 (0), given: 78

GMAT Tests User
Promotion
ngoctraiden1905 wrote:
I think it's easier: because all of the term divisible by 3 so the remainder of the multiply of them will be the remainder when after dividing by 3 then dividing by 11. This can only be 3 because all other choice is greater than 11


That's not right:

Consider this 6*6*6 divided by 33. What is remainder?

According to you logic: as all of them are divisible by 3 remainder must be less than 11, but in this case remainder is 18>11.

For the original question the answer IS 30. But my point was that the solution I provided is not easy, so I wonder if there is some easier way to do the same.

_________________
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 10:51 am 
Offline
VP
VP
User avatar

Joined: Tue Apr 01, 2008
Posts: 1013
Location: Bangalore, Baroda
Followers: 5

Kudos (?): 78 (0), given: 18

GMAT Tests User
To find the remainder, we need to first find what are the remainders when 1044, 1047, 1050 and 1053 are divided by 33 individually.

1044/33=31.......21
1047/33=31.......24
1050/33=31.......27
1053/33=31.......30

Now we need to find the remainders when 21*24 and 27*30 are divided by 33.

21*24/33=15......9
27*30/33=24......18

Finally we need to find the remainder when 9*18 is divided by 33:

9*18/33=4........30

So the answer is (C)


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 9:02 pm 
Offline
Manager
Manager

Joined: Sun Jul 05, 2009
Posts: 221
Followers: 0

Kudos (?): 2 (0), given: 5

GMAT Tests User
Economist wrote:
To find the remainder, we need to first find what are the remainders when 1044, 1047, 1050 and 1053 are divided by 33 individually.

1044/33=31.......21
1047/33=31.......24
1050/33=31.......27
1053/33=31.......30

Now we need to find the remainders when 21*24 and 27*30 are divided by 33.

21*24/33=15......9
27*30/33=24......18

Finally we need to find the remainder when 9*18 is divided by 33:

9*18/33=4........30

So the answer is (C)


Thanks, this is indeed a easier way. It also seems to work for any number :lol:

Still a bit long solution......


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sat Oct 24, 2009 10:46 pm 
Offline
Senior Manager
Senior Manager

Joined: Thu Nov 03, 2005
Posts: 424
Location: Chicago, IL
Followers: 0

Kudos (?): 7 (0), given: 16

GMAT Tests User
OA is C.

_________________
Hard work is the main determinant of success


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sun Oct 25, 2009 12:41 am 
Offline
Manager
Manager

Joined: Sun Jul 05, 2009
Posts: 221
Followers: 0

Kudos (?): 2 (0), given: 5

GMAT Tests User
rlevochkin wrote:
OA is C.


Any OE? Coz answer C is apparent but how to solve is the key here.


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Mon Oct 26, 2009 7:52 pm 
Offline
Intern
Intern

Joined: Mon Jul 13, 2009
Posts: 23
Followers: 0

Kudos (?): 1 (1), given: 1

Key to this remainder problem is the following properties,
you can multiply the remainders as long as you correct the excess remainders.

1044/33 yields R21
1047/33 yields R24 (Don't need to calculate as 1047 is 3 integers away from 1044)
1050/33 yields R27
1053/33 yields R30

Now, multiply remainders and correct the excess remainders.

R21* R24 * R27 * R30
Solve the problem till you get a remainder less than 33.

since 1044 and 33 are big numbers, it is possible to make arithmetic mistake here. Is there away to solve this problem
efficeintly?


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Tue Oct 27, 2009 12:49 am 
Offline
Manager
Manager

Joined: Tue Sep 15, 2009
Posts: 179
Followers: 0

Kudos (?): 3 (0), given: 1

calculate the remainder when 1044 is divided by 33 = 21

so the eqn becomes 21*24*27*30 / 33

= 27*27*21/33

considering negative remainders we get -6*-6*21/33

= 36*21/33

= 3*21/33

= 63/33

= 30.

I will go with option C


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Tue Oct 27, 2009 12:57 am 
Offline
Manager
Manager

Joined: Sun Jul 05, 2009
Posts: 221
Followers: 0

Kudos (?): 2 (0), given: 5

GMAT Tests User
deepakraam wrote:
calculate the remainder when 1044 is divided by 33 = 21

so the eqn becomes 21*24*27*30 / 33

= 27*27*21/33

considering negative remainders we get -6*-6*21/33

= 36*21/33

= 3*21/33

= 63/33

= 30.

I will go with option C


Wow!! Everyday I am learning new thing is this forum. Thanks.


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Tue Oct 27, 2009 9:54 pm 
Offline
Manager
Manager

Joined: Mon Oct 26, 2009
Posts: 59
Followers: 0

Kudos (?): 2 (0), given: 25

GMAT ToolKit User GMAT Tests User
useful insight..
thanks for the good question.


  Profile  
 
  Re: Remainders [#permalink]
PostPosted: Sun Nov 08, 2009 8:13 pm 
Offline
Intern
Intern

Joined: Sun Nov 08, 2009
Posts: 13
Followers: 0

Kudos (?): 3 (0), given: 12

smtripathi wrote:
I guess, we can prove this by taking two numbers in general format :

N1=dQ1+R1 ......1
N2=dQ2+R2 ......2

so, if we multiply them to find the remainder, it will become :
N1*N2=d^2*Q1Q2+dQ2*R1+dQ1*R2+R1*R2
or, =d[d*Q1*Q2+Q2*R1+Q1*R2] + R1*R2
so, we have remainder R1*R2 , but this may be bigger than d, so we need to rationalize R1*R2 to get the actual remainder.

I tend to remember things if I prove them .. so thought of adding my two cents.



good one . this can be extended to the SUM as well.

thanks


  Profile  
 
Online
gmatclubot
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 14 posts ] 


Who is online

Users browsing this forum: No registered users and 1 guest


Search for:
Jump to:




GMAT Club MBA Forum Home | Sitemap | About | Privacy Policy | Terms and Conditions | GMAT Club Rules | Contact
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group and phpBB SEO
Kindly note that GMAT (C) is a registered trademark of the Graduate Management Admission Council, and this site has neither been reviewed nor endorsed by GMAC.