last two digit of (123)^123!..explain

• ans: 01
  
  this is in the form 123^(x000000...)
  => 123^4*y000..
  =>z1^y00000..
  =>01.(because tens place=last digit of product of tens place from base and units
  place from power)
• 11 years agoHelpfull: Yes(17) No(1)
• 123! will end with more than two 0s.Hence it is divisible by 4.123!=4*x+0
  123^123!= 123^4 for last two digits divide 123 with 100==> 23^4=279841
  last two digits=41
• 11 years agoHelpfull: Yes(10) No(7)
• 123^123 or 123^123!.
• 11 years agoHelpfull: Yes(3) No(9)
• When the bases are same the powers will be added.. Now the question reduces to .
  123^7626( i.e. 1+2+3+.....123)
  Now find the last two digit by dividing it with 100(using remainder theorem)
  Most probably the answer will be 44
• 11 years agoHelpfull: Yes(2) No(6)
• for last two digits 01.
• 11 years agoHelpfull: Yes(1) No(0)
• answer is 27
  
• 10 years agoHelpfull: Yes(1) No(1)
• zero huh?
  coz 123 doesnt contain any zeroes ?
  
• 11 years agoHelpfull: Yes(0) No(6)
• 00 bcz 123! will contain some zeroes..and hence
• 11 years agoHelpfull: Yes(0) No(1)