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

• sum of digits of denominator 6+5+6+1=18 1+8=9
  sum of digits of numerator =90 9+0=9
  both nemerator nd denominator will be divisble by 9 without any remainder
  i.e. 0 remainder
• 10 years agoHelpfull: Yes(40) No(2)
• ans is 0
  as 123...789 is divisible by 3 and 6561=3^8
• 10 years agoHelpfull: Yes(4) No(3)
• zero is the answer.
  take unit place = (9)^24
  24/10= 2.4
  (9^2)^10.9^4=(81)^10.6561=1.6561/6561=0 remainder
• 10 years agoHelpfull: Yes(3) No(8)
• 0
  
  (3*K)24 %6561
  
  6561=3^8
  
  so,(3*K)24 will be completely divided by 3^8
• 10 years agoHelpfull: Yes(2) No(1)
• no.is multiple of 3 so
  (3*n)^24=((3^24)(n^24))/(3^8=6561)=0rem
  bcz (3^24)/(3^8) exectly
  
• 10 years agoHelpfull: Yes(2) No(0)
• soln is 0.
  6+5+6+1=18
  1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9=90
  90=9+0=9
  so 90/9=0 as remainder
• 9 years agoHelpfull: Yes(2) No(0)
• zero is the answer.
  take unit place = (9)^24
  24/10= 2.4
  (9^2)^3.9^4=(81)^3.6561=1.6561/6561=0 remainder
• 10 years agoHelpfull: Yes(0) No(12)
• Summation of all the digits is 90 thus the no. In the bracket can be witten as (9*k)^24/9^4 and hence the remainder will be zero.
  
• 10 years agoHelpfull: Yes(0) No(0)
• 0
• 10 years agoHelpfull: Yes(0) No(0)
• zero is the answer.
  take unit place = (9)^24
  24/10= 2.4
  (9^2)^10.9^4=(81)^10.6561=1.6561/6561=0 remainder
  
• 9 years agoHelpfull: Yes(0) No(0)
• (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 = 94. So,
  (924)/(94) Reminder is 0
• 3 years agoHelpfull: Yes(0) No(0)