Maximum value of N for {(100^8)-(111112^2)} is divisible by 3^N. What will be maximum value for N?
(a)6
(b)2
(c)1
(d)0

• 99888888*100111112 so 9+9+8+8+8+8+8+8=66 66 is divisible by 3 but not 9 so ans should be ->3^1
  
• 8 years agoHelpfull: Yes(15) No(2)
• Using a^2-b^2 we get this:-
  99888888*100111112
  Hence least significant bit have 6 i.e at first position which won't be divisible by any power of 3 except 3^0...
  So d) option i.e 0 is correct answer.
• 8 years agoHelpfull: Yes(9) No(11)
• Option C) N=1
  (100^4)^2-(111112)^2=100000000-111112=99888888
  which is divisible by only 3
  so N=1
• 8 years agoHelpfull: Yes(7) No(5)
• This would be (c)1.
  Using a^2-b^2 we get this:-
  
  (888888)(1111112)
  Hence only 3^1 would be divisible by this number
• 8 years agoHelpfull: Yes(3) No(2)
• Answer will be n=1 bcoz here the no will get divided by both 3 and 1(anything to the power 0 is 1). But the question here is max value of n. So max value will be n=1... Option c...
• 8 years agoHelpfull: Yes(2) No(0)
• Ans would be c) not d) sorry guyzz....
  Since 99888888 would be divisible by 3^1 only..
• 8 years agoHelpfull: Yes(1) No(0)
• i m also agree with kunal...ans should be c)1
  [(100)^4]^2-(111112)^2
  i.e a^2-b^2=(a+b)(a-b)
  98888888*100111112,,,,at first position multiple will be 6 which will be divisible by 3..
  
• 8 years agoHelpfull: Yes(1) No(1)
• Hello Manish, Would u plz explain how you can say that if a no. has 6 at ones position then it wont be divisible by any power of 3 except 0?
• 8 years agoHelpfull: Yes(0) No(1)