Find number of zeros after decimal point in (5/6)^40?
(Explanation requested)

• (5/6)^40
  
  let us find the remainder first i.e:
  
  (-1)^40 = 1
  
  hence (5/6)^40 can be written as (6^40)*m + 1 where m is any positive integer
  
  so the remainder will create the decimal structure hence the number after decimal will be 1/(6^40)
  
  now the number of digits in 6^40 is = floor(1+log(6^40))=32
  
  now when 1 is divided by a 32 digit number it will have 31 zeroes after decimal point for example....
  
  1/2=0.5-----------0 zero
  1/22=0.04---------1 zero
  1/222=0.004-------2 zeroes
  
  hence when 1/(32 digits)---------------31 zeroes
  
  hence the ANS-------31
• 8 years agoHelpfull: Yes(0) No(0)