sign up I login
 advanced
refer a friend - earn nickels!!

Ask & Discuss Questions with Community & Experts

 Moderation Team
Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: The remainder when x=1! + 2! + 3! + 4!+ .....+ 100! is divided by 240 is
Forum Index -> Algebra -> View Full Question like the article? email it to a friend.  
Author Message
[Post New]13 May 2008 16:22:57 IST
    Subject: Re:The remainder when x=1! + 2! + 3! + 4!+ .....+ 100! is divided by 240 is
[Up]
allamraju (3422)

Blazing goIITian

Olaaa!! Perrrfect answer. 606  [802 rates]
allamraju's Avatar
total posts: 1014    
offline Offline
Ans is 153,bcoz 1!+2!+3!+4!+5!+720+......=1+2+6+24+120+240k where k is a constant=153+240k.Hence the ans.
MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC.
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
You have to be logged on to rate
  
 

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya