1!+2!+3!...+50! Divided by 5! Remainder will be ?

1) 0
2) 33
3) 77
4) 1

• remainder will be same as remainder obtained when 1!+2!+3!+4! is Divided by 5!
  
  so 1!+2!+3!+4! = 1+2+6+24 = 33
  so 33 is the remainder when 1!+2!+3!...+50! Divided by 5!.
• 11 years agoHelpfull: Yes(21) No(7)
• Answer is 0.
  
  1+2+3+4+5 = 15/5, remainder is zero.
  
  Then after sum of next 5 numbers in the series are divisible by 5 with zero remainder.
• 11 years agoHelpfull: Yes(4) No(17)
• 1+2+6+24=33 ans
• 11 years agoHelpfull: Yes(2) No(0)
• For any natural number m>=5, m! is divisible by 5!. So remainder is 1!+!+3!+4!when
  1!+2!+.....50! is divided by 1.
• 11 years agoHelpfull: Yes(1) No(4)
• 33 .. 1! +2! +3! +4! +5!(1+6+7*6....)
  1!+2!+3!+4! =33 and remaining terms are cancelled by 5! .. so remainder is 33
• 11 years agoHelpfull: Yes(1) No(0)
• Remainder is 80
• 11 years agoHelpfull: Yes(0) No(6)
• 0 n(n+1)/2 to get sum of 1 to 50=1275 this ans divisible by 5 so remainder will be "0"
• 11 years agoHelpfull: Yes(0) No(5)
• the ans will be 33
• 11 years agoHelpfull: Yes(0) No(0)
• !+2!+3!+.............+50!
  1! is the smallest and 50! is the highest
  now the half of 50! is 25! and if we increase 1! in 50! it will give 51!
  Now 25! * 51!=1275!
  So,1!+2!+3!+...........+50!=1275!
  now divide 1275! with 5!=255 and 1275 is completely divisible by 5 .
  Hence, (1!+2!+3!+.................+50!)/5 then the remainder will be zero because it is completely divisible by 5.
• 11 years agoHelpfull: Yes(0) No(1)