(1!^1!+2!^2!+3!^3!+4!^4!+......+50!^50!)/5
The remainder for the above is--------?

• remainder for above is 2.
  no. ahead 5!^5 will be completely divisable by 5.
  rest 1!*1+2!^2+3!^3+4!^4 = 331997
  when divided by 5 gives remailder 2. ans :)
• 12 years agoHelpfull: Yes(24) No(3)
• 5! ^ 5! and above terms will be divisible by 5.
  so take the terms before that.
  1 + 2^2 + 6^6 + 24^24 /5 (check for the remainders , dont calculate this.)
  4+1+1+1/5 (1 leaves 4 as remainder , 2^2 leaves 1 , 6 wil always giv 1 , 24^24 wil also give 1 as 4 as unit digits in even powers as 6)
  7/5 leaves 2 as remainder.
• 12 years agoHelpfull: Yes(11) No(3)
• the anwser is 2
  
• 12 years agoHelpfull: Yes(1) No(2)
• @Mr.prasanth, 1 leaves 1 as remainder not 4
• 12 years agoHelpfull: Yes(0) No(2)
• ater 5!^5! the remainder will be 0 so,
  (1^1!+2^2!+6^3!+24^4!)/5
  no calculate for rtemainder as for
  1st term remainder=1
  2nd term remainder=-1
  3rd term remainder=1
  4th term remainder=-1^(4!)=1
  so answer =2
• 12 years agoHelpfull: Yes(0) No(2)
• after 5!^5! the remainder will be 0 so,
  (1^1!+2^2!+6^3!+24^4!)/5=331997/5
  so remainder =2
• 11 years agoHelpfull: Yes(0) No(0)