1!+2!+3!...+50! divided by 5!
remainder will be ?

• 1!+2!+3!...+50! divided by 5!
  remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
  
  so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).
  
  so remainder is 33.
• 11 years agoHelpfull: Yes(44) No(7)
• Clearly 5! onwards each is divisible by 5!.
  SO the remainder will come from the division (1!+2!+3!+4!) divided by 5!
  Now (1!+2!+3!+4!) = 1+2+6+24 = 33 and 5! = 120 so the remainder is 33.
• 11 years agoHelpfull: Yes(19) No(2)
• 33 is correct.
  
  when 33 is divided by 5! ( 120) , u will get 33 as remainder.
  
  @ Vivek,
  I think u r considering the case when 33 is divided by 5 .
• 11 years agoHelpfull: Yes(6) No(2)
• @ amar 33 is not possible dude it will be 3.
• 11 years agoHelpfull: Yes(4) No(30)
• answer will be zero.........as the total for this sum is 3.1035x10^64 which is easily divisible by 5
• 11 years agoHelpfull: Yes(3) No(15)
• till 4! last digit will be non zero.
  after 4! all terms will have last digit as 0 so it doesn't matter how many terms are there last digit is always zero and sum is divisible by 5.
  Now, if we see 1!+...+4!=33
  Therefore last digit of 1!+......+50! will be 3
  hence remainder is 3.
• 11 years agoHelpfull: Yes(2) No(1)
• let us see it in simple way here we are taking 33/120 i.e. (1!+2!+3!+4!)/5!
  now, take an example divide(12/6) is cancelled by (2/1) similarly when we cancel 5 with any digit it get cancelled if digit is greater than 5.
• 11 years agoHelpfull: Yes(1) No(4)
• 1!+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(1) No(5)
• 1!+2!+3!...+50! divided by 5!
  remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
  
  
  
  so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).
  
  so remainder is 33
  
  can anyone explain me this what to divide with whom?
• 7 years agoHelpfull: Yes(1) No(0)
• pls can anyone explain me the complete solution. concept behind taking only 1! to 4!
• 11 years agoHelpfull: Yes(0) No(0)
• 1!+2!+3!...+50! divided by 5!
  remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
  
  so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).
  
  so remainder is 33
  
• 8 years agoHelpfull: Yes(0) No(0)
• The remainder when the terms greater than 5! are divided by 5! becomes 0 so we need to consider the terms upto 4!.
  So remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
  So remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120)
  So remainder is 33.
• 7 years agoHelpfull: Yes(0) No(0)
• Reminder 3
• 6 years agoHelpfull: Yes(0) No(0)
• 1! + 2! +3! +4! +5! ......+50!
  after 5! eery term will be divisible by 5! beacuse 5! is the factor
  now , 1 + 2+6+24=33
  33 will be the reminder
• 4 years agoHelpfull: Yes(0) No(0)