Latest TCS Aptitude Question SOLUTION: Find the remainder when 32^33^34 is divided by 11.

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(#M40028006) TCS Numerical Ability question tcs Keep an EYE

Find the remainder when 32^33^34 is divided by 11.

Asked In TCS rakesh (10 years ago)

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shankar kumar  (10 years ago)

answer=1
32^33^34= (33-1)^33^34/11= (-1)^33^34/11= (-1)^even number(33*34) /11
= 1/11= 1(remainder)

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ya  (10 years ago)

answer=10
32^33^34= (33-1)^33^34/11= (-1)^33^34/11= (-1)^odd no.(as last digit of 33*34 is odd)
= -1/11= -1 or 11-1=10(remainder)

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Kundan Kumar  (10 years ago)

32^33^34/11
can be written as
32^N/11 where N= 33^34
Now, term reduces to 10^N/11
Now find the cyclicity:-
10^1/11 = 10 ( remainder)
10^2/11= 1 remainder)
10^3/11 = 10 ( remainder)
10^4/11= 1 (Remainder)
Hence it has cyclicity of 2
now question reduces to
N/2
Now find the remainder
of 33^34/2
Remainder= 1
First term of the cycle will be the answer
Hence Final answer is 10

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Rachit  (10 years ago)

10
(33-1)^odd number(as odd^even is always equal to odd)/11
==-1/11
= 10

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yaswant kumar chaudhary  (10 years ago)

answer will be 10 .

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yaswant kumar chaudhary  (10 years ago)

ans 0.
32^33^34/11 write it as 32^n/11 remainder in this case is calculated first.it can be written as 10^n/11 or (-1)^n/11 so remainder will be -1 or 10.its cycle will be 2.now 33^34/2 gives 1 remainder so the whole remainder will be 0.

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Guru Pranav  (10 years ago)

10 is ans.

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