999^x-999^y when divided by 1000 will give some remainder.How many such remainders will be possible?
options:
a) 3
b) 8
c) 99
d) more than 99

• we can write the equation like this
  [(999^x)/1000]-[(999^y)/1000]
  which is a form of [(a^x)/(a+1)]-[(a^y)/(a+1)]
  and we know [(a^n)/(a+1)] gives remainder "1" whn n is even and gives remainder "a" when n is odd.by taking even odd combination of x and y only 2 remainder are possible .so ans is 3(as question was including zero).
• 11 years agoHelpfull: Yes(17) No(0)
• A)3
  999^x/1000=(-1)^x
  Similarly 999^y=(-1)^y
  hence we get [(-1)^x]+[(-1)^y] which can result in 3 combinations-0,2,-2
• 11 years agoHelpfull: Yes(4) No(4)
• 0,7 and 8 are possible solutions.
  so answer is 3.
• 11 years agoHelpfull: Yes(1) No(1)
• 999^x - 999^y = (1000-1)^x - (1000-1)^y
  Depending on the value of X and Y being even or odd we will get
  
  x = even & y = even remainder = 0
  x = odd & y = even remainder = -2
  x = even & y = odd remainder = 2
  x = odd & y = odd remainder = 0
  
  so there can be total 3 possible remainders
• 11 years agoHelpfull: Yes(1) No(1)
• given x>y>0
• 11 years agoHelpfull: Yes(0) No(3)