find the sum of 1!+2!+3!+4!+...100!

• abe barnisa ki bacchi toda chota solution nhi de sakti!
• 11 years agoHelpfull: Yes(17) No(0)
• 101!/100 - 100!(1/100+2/100...+1), so 100!(1+1/100)=100!*101/100=101!/100
• 11 years agoHelpfull: Yes(10) No(10)
• This looks more like a programming question than an aptitude question to me!!
  Well, The solution can be derived as:
  Summation(1!+2!+3!+....+100!)
  Summation(n) where n=1 to n=100
  The answer: 94269001683709979260859834124473539872070722613982672442938359305624678223479506023400294093599136466986609124347432647622826870038220556442336528920420940313
  
  The above is a computer simulated result!!
• 11 years agoHelpfull: Yes(8) No(40)
• let us take 100! common from the series:
  100!( 1!/100!+ 2!/100!+ 3!/100!+....+ 99!/100! + 100!/100!)
  most of the terms will be negligible..
  100!(1/100 + 1)
  101*100!/100
  101!/100 Ans
• 11 years agoHelpfull: Yes(6) No(0)
• @barnisa :
  lol!
• 11 years agoHelpfull: Yes(3) No(1)
• @payal & @saloni Kya karu yaar!! Iss jaise question ka toh aisa hi jawab hona chahiye naah...
• 11 years agoHelpfull: Yes(2) No(0)
• kya bat chat0
  
• 11 years agoHelpfull: Yes(1) No(1)
• (n+1)!/n is formula
• 11 years agoHelpfull: Yes(1) No(0)
• la jawaab!
• 11 years agoHelpfull: Yes(0) No(0)
• give the options. last digit would be 3. then check from opsns
• 11 years agoHelpfull: Yes(0) No(0)
• 101!-100! = 101(100!)-100! = 100!(101-1) = 100! * 100
• 11 years agoHelpfull: Yes(0) No(1)
• may be ans is 101!-1
• 11 years agoHelpfull: Yes(0) No(3)
• Barnisa ki pan kore ans ta diacisss...?
• 11 years agoHelpfull: Yes(0) No(0)
• 5 will be the answer.
  after 7! the remainder will be zero.
  so 1!+2!+3!=4!+5!+6!=873 divide by 7 we have
  5 as remainder
• 11 years agoHelpfull: Yes(0) No(3)
• 101!/100 will be the sum !!!!
• 11 years agoHelpfull: Yes(0) No(0)