Latest TCS Aptitude Question SOLUTION: {(29^31)^109}/9

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(#M40165034) TCS Numerical Ability question Find the remainder Keep an EYE

{(29^31)^109}/9

Asked In TCS karuppusamy (6 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

vinod  (6 years ago)

step 1st :
take (31^109)=N
step 2nd:
solve (29^N)/9
=on which 2 get reminder on dividing 29/9 so,
(2^N)/9
step 3rd:
now find the power cycle when remainder became 1 as
(2^1)/9=2(remainder)
(2^2)/9=4(remainder)
(2^3)/9=8/9= -1 or 8
so remainder -1 get on 3rd cycle so 1 get in 6th cycle:
step 4:
using 6 as power cycle we divide N
p== N/6=(31^109)/6=(1^109)/6 as 31/6=1 as per reminder theorem
step 5:
now put p in step 2 as value of N..
{(29^31)^109}/9==(2^p)/9
=(2^1)/9 as p=1
=2 as per reminder theorem
so final remainder is 2..

Is this solution Helpfull? Yes (7) | No   

Sanjana Reddy B  (6 years ago)

(29^31) => 9^1(odd number)=9
(29^31)^109 => 9^9(odd number)=9
(29^31)^109/9 => 9/9=1;
thus the reminder becomes zero

Is this solution Helpfull? Yes (2) | No (7)  

Sunkara Silpa  (6 years ago)

First step 29/9 =reminder is 2 geting so {(2^31)^109}/9
Second step =>31*109= 3379 => (2^3379)/9
Third step=> 2^6/9 gives the reminder 1 so divide 3379/6 we will get reminder as 1(Here 6 is as 2^1/9=2, 2^2/9=4 like power cycle is 6)

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