Latest TCS Aptitude Question SOLUTION: Find the remainder 2^2014^302 when divided by 9.....a star mark question

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

Asked In TCS Dipika Jha (10 years ago)

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Kundan Kumar  (10 years ago)

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)

Is this solution Helpfull? Yes (40) | No (7)  

palash verma  (10 years ago)

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

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Sachin Grover  (10 years ago)

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

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