when (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! is divided by 5,the remainder obtained is?

a. 2
b. 0
c. 4
d. none of these

• a) 2
  (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! / 5
  
  = 1 + 2^2 + 6^6 + 24^24 + [(5!)^5! + (6!)^6! + .....+ (100!)^100!] / 5
  
  from (5!)^5! to 100!^100! each term is divisible by 5 & gives rem = 0
  
  reqd rem = 1 + 2^2 + 6^6 + 24^24 / 5 => rem= 7/5 = 2
  [unit digit of 1 + 2^2 + 6^6 + 24^24 = 1+4 +6+6 = 7]
  
  
• 10 years agoHelpfull: Yes(91) No(0)
• 1 4 6^6 24^24 120^120....
  from 120 the last digit is 0
  for 6^6 last digit is 6 and for 24^24 odd powers will give 4 and even powers will give 6
  so 1+4+6+6=17 gives remainder 2 when divided by 5
  ans is 2
• 10 years agoHelpfull: Yes(6) No(0)
• ANSWER==> d)none of this
  as (1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! = 1+4+6^6+....
  the last digit of the summation will be 1. so the remainder will be 1
• 10 years agoHelpfull: Yes(2) No(7)
• the ans is (d)
• 10 years agoHelpfull: Yes(2) No(6)
• (5!+.......100!)/5, rem is 0.
  then 1!^1!=1,2!^2!=4,3!^3!=6,4!^4!=6
  then (1+4+6+6)/5
  =17/5
  =2
  ans:a)2
  
• 10 years agoHelpfull: Yes(2) No(0)
• last digit will be zero
  so it is divisible by 5
  rem=0
• 10 years agoHelpfull: Yes(1) No(13)
• option b 0
• 10 years agoHelpfull: Yes(0) No(5)
• @Rakesh: thanks for correcting me!
• 10 years agoHelpfull: Yes(0) No(0)
• remainder 0 , since last digit is zero
• 10 years agoHelpfull: Yes(0) No(5)
• after 5! oll the no. will be divisible by 5., as they contain 5 in every term., so add up 1!+2!..+4!/5 ans.(2)
  
• 10 years agoHelpfull: Yes(0) No(0)
• plz send any of 1 tcs qustn pepr...
• 10 years agoHelpfull: Yes(0) No(0)
• CAN ANY ONE SEND ME THE TCS PAPER TO MA MAIL ID IS pampanagouda032@gmail.com
• 10 years agoHelpfull: Yes(0) No(0)
• 1!(mod 5) = 1 (mod 5)
  2!(mod 5) = 2 (mod 5)
  3!(mod 5) = 6 (mod 5) =1(mod 5)
  4!(mod 5) = 24 (mod 5)= 4 (mod 5)
  5!(mod 5) = 120 (mod 5)= 0(mod 5)
  when n>=5,then mod value become 0.
  hence, (1!+2!+3!+.......+100!) (mod 5) = (1+2+1+4+0+0+.....+0)(mod 5)
  =8(mod 5) =3(mod 5)
  so,required remainder is 3...
  ANS:d. -----> none of these...
• 7 years agoHelpfull: Yes(0) No(0)
• i do not under stand one thing.... why more then one people giving same explanation
• 5 years agoHelpfull: Yes(0) No(0)