What will be the reminder when (1234567890123456789)^24 is divided by 6561 ?

• The digital sum of number is
  (1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9)=90=(9+0)=9
  So it is divisible by 3 or 9
  The given divisor 6561 = 9^4
  So
  9^24/9^4 Reminder is 0
  Ans : 0
• 8 years agoHelpfull: Yes(27) No(1)
• sum of no. is 90 so its divisible by 3 0r 9. Similarly 6561 is also divisible by 3 or 9. S Remainder is 0.
• 8 years agoHelpfull: Yes(10) No(4)
• total sum 90 so (9+0=9) its factor is 3 or 9
  then 6561=9^4
  therefore 9^24/9^4 and their remainder is 0
  so answer is 0
• 8 years agoHelpfull: Yes(2) No(0)
• Sorry I did mistake.
  Ans is 2*2*2*2*2*1*1*1*1 =32
• 8 years agoHelpfull: Yes(1) No(10)
• 0
  9^4=6561
  then (9^4)^6
• 7 years agoHelpfull: Yes(0) No(0)
• {(9*x)^n}/(9^m) always leaves the reminder 0 if n>m and if n
• 5 years agoHelpfull: Yes(0) No(0)