(100)^8-(1112)^2 is divisible by 3^n, then what is the maximum value of n.
(a) 6
(b) 2
(c) 1
(d) 0

• If v will solve equation after substraction v will get = 999999998763456
  And if v will add all the digits v will get 111
  So it is divisible by 3 only
  Therefore, ans is 1
  
  Ans--> C
• 9 years agoHelpfull: Yes(8) No(2)
• 100...16* times -1236544=9999999998763456.
  Sum of the digits is 120. Which is only divisible by 3^1
  Ans→1
• 9 years agoHelpfull: Yes(4) No(0)
• a^2 - b^2 = (a+b)(a-b)
  =((10^4)^2)-(1112^2)
  = (10^4 + 1112)(10^4 - 1112)
  =(10000+1112)(10000-1112)
  =(11112)(8888)
  =98763456
  ANS---->c
• 9 years agoHelpfull: Yes(4) No(0)
• a^2 - b^2 = (a+b)(a-b)
  = (10^8 + 1112)(10^8 - 1112)
  try for first bracket as it will be easier,it will be divided by 3 only.
• 9 years agoHelpfull: Yes(1) No(1)