What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

It is currently 05 Dec 2016, 04:42
GMAT Club Tests

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11

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

Hide Tags

5 KUDOS received
Manager
Manager
avatar
Joined: 19 Aug 2009
Posts: 79
Followers: 1

Kudos [?]: 218 [5] , given: 46

What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

Show Tags

New post 04 Nov 2009, 00:07
5
This post received
KUDOS
14
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

72% (10:39) correct 28% (01:49) wrong based on 248 sessions

HideShow timer Statistics

What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?

1. 125
2. 225
3. 325
[Reveal] Spoiler: OA
Expert Post
10 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 35867
Followers: 6837

Kudos [?]: 89914 [10] , given: 10382

Re: Numbers 3 [#permalink]

Show Tags

New post 31 Jan 2010, 06:17
10
This post received
KUDOS
Expert's post
9
This post was
BOOKMARKED
jeeteshsingh wrote:
Is there a specific approach to tackle this?


No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

3 KUDOS received
Intern
Intern
avatar
Affiliations: CA - India
Joined: 27 Oct 2009
Posts: 45
Location: India
Schools: ISB - Hyderabad, NSU - Singapore
Followers: 20

Kudos [?]: 681 [3] , given: 5

Re: Numbers 3 [#permalink]

Show Tags

New post 04 Nov 2009, 03:19
3
This post received
KUDOS
The ans has to be 225. all the terms in the sequence after (3!)^3 are divisible by 1152 and hence remainder is 0. Upto (3!)^3, sum of all numbers, i.e. 1+8+216 = 225 which is the remainder!!
Senior Manager
Senior Manager
User avatar
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 368 [0], given: 47

GMAT ToolKit User
Re: Numbers 3 [#permalink]

Show Tags

New post 31 Jan 2010, 05:01
Is there a specific approach to tackle this?
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!! :beer

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|


~~Better Burn Out... Than Fade Away~~

Manager
Manager
avatar
Joined: 10 Feb 2010
Posts: 193
Followers: 2

Kudos [?]: 103 [0], given: 6

Re: Numbers 3 [#permalink]

Show Tags

New post 12 Feb 2010, 17:45
Nice Explanation!
Manager
Manager
avatar
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 50 [0], given: 16

Re: Numbers 3 [#permalink]

Show Tags

New post 14 Jul 2011, 02:13
Bunuel....+1 to you...
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat ;)

Satyameva Jayate - Truth alone triumphs

Manager
Manager
User avatar
Joined: 14 Apr 2011
Posts: 199
Followers: 2

Kudos [?]: 23 [0], given: 19

Reviews Badge
Re: Numbers 3 [#permalink]

Show Tags

New post 16 Jul 2011, 02:13
good question. Thanks Bunuel for sharing the approach for these problem!
_________________

Looking for Kudos :)

Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 40 [0], given: 11

Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

Show Tags

New post 19 Oct 2012, 03:21
Did .. the same thing as Bunuel..

But took 8 minutes..

In actual exam.. I would have guessed and moved on

What is the source of this problem?
Manager
Manager
avatar
Joined: 14 Nov 2011
Posts: 149
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 12 [0], given: 103

GMAT ToolKit User
Re: Numbers 3 [#permalink]

Show Tags

New post 18 Jul 2013, 05:20
Bunuel wrote:
jeeteshsingh wrote:
Is there a specific approach to tackle this?


No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.


Hi Bunnel,
To get the remainder, we dont have to reduce the fraction right?
That is we cant do - 225/1152 = 25/ 128 and get remainder 25?
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 35867
Followers: 6837

Kudos [?]: 89914 [0], given: 10382

Re: Numbers 3 [#permalink]

Show Tags

New post 21 Jul 2013, 01:24
cumulonimbus wrote:
Bunuel wrote:
jeeteshsingh wrote:
Is there a specific approach to tackle this?


No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.


Hi Bunnel,
To get the remainder, we dont have to reduce the fraction right?
That is we cant do - 225/1152 = 25/ 128 and get remainder 25?


Yes, 225 divided by 1152 yields the remainder of 225. The same way as 2 divided by 4 yields the remainder of 2, not 1 (1:2).
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 12871
Followers: 559

Kudos [?]: 157 [0], given: 0

Premium Member
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

Show Tags

New post 08 Aug 2014, 05:49
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Director
Director
User avatar
Joined: 23 Jan 2013
Posts: 567
Schools: Cambridge'16
Followers: 1

Kudos [?]: 41 [0], given: 40

CAT Tests
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

Show Tags

New post 12 Aug 2015, 11:22
starting from 4!^3=24^3 all numbers are divisible by 1152, i.e. remainder equal to 0

Only 1!^3, 2!^3 and 3!^3 are not divisible by 1152 and have remainder equal to 1,8 and 216, respectively.

If sum numbers we can sum their remanders to find total remainder, which is 216+8+1+0=225

B
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 12871
Followers: 559

Kudos [?]: 157 [0], given: 0

Premium Member
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

Show Tags

New post 26 Aug 2016, 22:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11   [#permalink] 26 Aug 2016, 22:31
    Similar topics Author Replies Last post
Similar
Topics:
1 If Z = 11 + 22 + 33 +...+ 88, what is the remainder when Z is divided stonecold 1 26 Oct 2016, 03:20
1 Experts publish their posts in the topic What will be the remainder when 13^36 is divided by 2196? aayushagrawal 1 08 Jun 2016, 18:39
17 Experts publish their posts in the topic What is the remainder when 1555 * 1557 * 1559 is divided by 13? Bunuel 8 16 Apr 2015, 04:03
25 Experts publish their posts in the topic What is the remainder when (1!)!^3 + (2!)^3 + (3!)^3 + … + ( langtuprovn2007 6 23 Jun 2014, 15:44
Experts publish their posts in the topic What is the remainder when 43^43+ 33^33 is divided by 10? rashi22 6 03 Sep 2010, 23:15
Display posts from previous: Sort by

What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11

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


GMAT Club MBA Forum Home| About| 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®.