Did anyone solve the question which was like

Q. z=y^o+y^1+.....+Y^n
and if the equation is divided by 6...dont remember the question exactly...

• always divisible by 2 is the answer
• 11 years agoHelpfull: Yes(11) No(0)
• divisible by 2
  the series is in g.p
  so a=9 r=9
  formula= a(r^n-1)/r-1
  so (9*9^n-1/8)/6
  if you take n value is 2
  then
  9*80/48=15 here reminder is 0
  so always divisible by 2 only
  
• 11 years agoHelpfull: Yes(7) No(0)
• Z=9^1+9^2+9^3+.....9^n if z is divided by 6 then remainder will be 0. what can be said about n?
  a) n is always divisible by 9
  b) n is always divisible by 3
  c) n is always divisible by 6
  d) n is always divisible by 2
  I think this was the question ...
• 11 years agoHelpfull: Yes(5) No(0)
• n always div by 2
• 11 years agoHelpfull: Yes(2) No(2)
• answer is always divisible by 3
• 11 years agoHelpfull: Yes(2) No(2)
• if u divide 9^1,9^2......,the remainder is always 0
  se answer is 3
• 11 years agoHelpfull: Yes(1) No(4)
• always divisible by 2
  
• 11 years agoHelpfull: Yes(0) No(1)
• alwayz divisible by2
  z=9(9^n-1)/8
  z is divisble by 6 ie 9(9^n-1)/8*6
  3(9^n-1)/8*2
  3(8n+1n-1)/8*2 using binomial exp--(x+1)^n=xn=1^n
  3(8n)/8*2=3n/2 therefore n shouble be divisible by 2
• 10 years agoHelpfull: Yes(0) No(0)