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

 It is currently 01 Jun 2020, 07:46 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11

Author Message
TAGS:

### Hide Tags

Manager  Joined: 19 Aug 2009
Posts: 77
What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11  [#permalink]

### Show Tags

6
35 00:00

Difficulty:   45% (medium)

Question Stats: 72% (02:18) correct 28% (02:12) wrong based on 267 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
Math Expert V
Joined: 02 Sep 2009
Posts: 64158

### Show Tags

19
6
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.
_________________
Intern  Affiliations: CA - India
Joined: 27 Oct 2009
Posts: 33
Location: India
Schools: ISB - Hyderabad, NSU - Singapore

### Show Tags

5
1
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!!
##### General Discussion
Manager  Joined: 22 Dec 2009
Posts: 225

### Show Tags

1
Is there a specific approach to tackle this?
Manager  Joined: 10 Feb 2010
Posts: 116

### Show Tags

Nice Explanation!
Manager  Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 150
Schools: Johnson '15

### Show Tags

Bunuel....+1 to you...
Manager  Joined: 14 Apr 2011
Posts: 157

### Show Tags

good question. Thanks Bunuel for sharing the approach for these problem!
Intern  Joined: 23 May 2012
Posts: 28
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11  [#permalink]

### Show Tags

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  Joined: 14 Nov 2011
Posts: 113
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 64158

### Show Tags

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).
_________________
Director  G
Joined: 23 Jan 2013
Posts: 507
Schools: Cambridge'16
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11  [#permalink]

### Show Tags

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
Manager  S
Joined: 06 Jun 2013
Posts: 143
Location: India
Concentration: Finance, Economics
Schools: Tuck
GMAT 1: 640 Q49 V30
GPA: 3.6
WE: Engineering (Computer Software)
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11  [#permalink]

### Show Tags

factorize 1152 into 2^7*3^2

see carefully all are cubic factorial

we need to look beyonf (4!)^3 as from here onwards remainder is zero

so consider cubic factorial of 1 , 2 and 3 and sum will give 225. and this number on division by 1152 gives 225.

Non-Human User Joined: 09 Sep 2013
Posts: 15033
Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11  [#permalink]

### Show Tags

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.
_________________ Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11   [#permalink] 21 Mar 2020, 05:13

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