what is the remainder when (1234567890123456789)^24 divided by 6561

• ans is 0
  as 123...789 is divisible by 3 and 6561=3^8
• 11 years agoHelpfull: Yes(26) No(3)
• (1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9) = 90
  
  and 6561 = 3^8
  
  so 90 is divisible by 3 thrfor rem will be 0
• 11 years agoHelpfull: Yes(17) No(2)
• The given numbr (123..........9)^24 =(3*x)^24 ,where x =(123....9)/3
  =(3^24)*(x^24)
  =(3^(8*3))*A, where A=x*24
  =(3^8)*(3^8)*(3^8)*A
  =(3^8)*B ,where B=(3^16)*A
  =6561*B ,(since 6561=3^8)
  Thus ,the given numbr(123....9)^24 is divisible by 6
• 11 years agoHelpfull: Yes(4) No(2)
• It is 0 as both are divisible by 3 where 6561=3^8.
• 11 years agoHelpfull: Yes(0) No(2)
• 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
• 11 years agoHelpfull: Yes(0) No(0)
• It is divisible by 6561.
• 11 years agoHelpfull: Yes(0) No(0)