consider p=(999^x-999^y) where x>y>0 are the natural numbers , what are possible remainders when divisible with 1000 where remainders (including 0)

• Ans- 0 , 2, 998
  depends on x,y
  if both(x,y) are even or odd then remainder is : 0
  if x is odd and y is even then rem : 998
  if x is even and y is odd then rem : 2
• 9 years agoHelpfull: Yes(12) No(0)
• (a)^n/(a+1) leaves remainder of
  a if n is odd
  else
  1 if n is even
  therefore there are 3 possible remainder i.e 0,2,998
• 9 years agoHelpfull: Yes(5) No(0)
• When both x,y are odd or even then remainder 0
  When x even and y odd then 1-(-1)= 2
  When x odd and y even then -1-(1)= -2
  ie; - ve remainder so again divide -2 by 1000 hence we get 998
  So 0,2,998 ans
• 9 years agoHelpfull: Yes(1) No(0)
• Find the total number of multiples of 3......as he said odd numbers...it will be
  66..
  Now we have remove the even numbers which are 108,114,120....total number of these will be 33
  Now..66-33=33.ans
• 7 years agoHelpfull: Yes(1) No(1)
• ans: 2
  apply remainder theorm
• 9 years agoHelpfull: Yes(0) No(0)
• can anyone explain the answer 2?? how answer could be 2??
  
• 9 years agoHelpfull: Yes(0) No(0)