Find the remainder 2^2014^302 when divided by 9.....a star mark question

• Firstly , Let N= 2014^302
  Now the question reduces to the :- 2^N /9
  Now proceed further by finding the cyclicity of 2 ( in Remainders)
  ie. divide 2^1,2,3,4 etc.. by 9 and write the remainder
  2^1 /9= 2(remainder)
  2^2 /9= 4(remainder)
  2^3 /9= 8(remainder)
  2^4 /9= 7(remainder)
  2^5 /9= 5(remainder)
  2^6 /9= 1(remainder)
  2^7 /9 = 2(reaminder)
  We can see that cyclicity is of 6( remainder repeat at an interval of 6)
  Now
  calculate the remainder when N/6 here N= 2014^302
  which will give the remainder is 4
  Now, answer will be the fourth term of the cyclicity of remainder 2
  Final answer is 7
  (Analyse the solution first)
• 11 years agoHelpfull: Yes(40) No(7)
• check cyclicity of remainder with diff power of 2..it will come as 6
  divide 2014 by 6..remainder will be 4..which means tht the 4th term in cyclicity is our remainder if 2^2014 is divided by 9 which is 7 (refer table above)
  now que remains 7^302 divided by 9
  again check cyclicity this time it will be 3
  302/3 gives rem as 2 and 2nd term in cyclicity gives rem as 4
  therefore 4 is our ans..
  the method is same as done above but have to be done in proper order
• 11 years agoHelpfull: Yes(25) No(3)
• Since 2^1=2, 2^2=4, 2^3=8, 2^4=16 ,2^5=32, 2^6=64, 2^7=128
  So, when the above values are divided by 9 then the cyclicity is found out to be 6. therefore,
  [(2^6)^335^304]*2^4/9 gives 16/9 i.e. 7 as remainder
• 11 years agoHelpfull: Yes(8) No(3)