BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Remainder

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members
Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 9:45 pm
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 261
Joined: Wed Mar 31, 2010 8:37 pmFollowed by: 3 members
Location: Varanasi
Thanked: 11 times

by ankur.agrawal » Sun Jan 16, 2011 11:43 pm
anshumishra wrote:
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
Absolutely Brilliant.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 857
Joined: Wed Aug 25, 2010 1:36 amFollowed by: 15 members
Thanked: 56 times

by AIM GMAT » Mon Jan 17, 2011 2:27 am
Wow !!! Nice explanation anshumishra . Thanks a lot for a nice method to solve the dangerous looking question.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Fri Jan 07, 2011 9:17 pm

by nehatandon » Mon Jan 17, 2011 5:56 am
anshumishra wrote:
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
wow, you made that look so easy. Excellent ! Thanks.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members
Thanked: 147 times

by anshumishra » Mon Jan 17, 2011 6:19 am
Thanks AIM GMAT, ankur, Neha !
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1