Latest TCS Aptitude Question SOLUTION: remainder of 3^1000 divided by 73

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remainder of 3^1000 divided by 73

Asked In TCS sathiya (7 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Nidhi  (7 years ago)

3^1 / 73 results in remainder of 3
3^2 / 73 results in remainder of 9
3^3 / 73 results in remainder of 27
3^4 / 73 results in remainder of 8

continuing same way we get a sequence of 3,9,27,8,24..... at 3^13/73 we have remainder equals to 3
i.e the sequence repeats at cycle of 12

hence 1000/12 gives a remainder 4
remainder 4 here implies to the 4th term of the sequence which is 8.
hence 8 is the required solution

Is this solution Helpfull? Yes (10) | No   

Priyaj  (7 years ago)

ans =8
3^1000/73=
=(3^4)250/73
=81^250/73
=8^250/73(remainder theorem)
=(8^3)^83 *8/73
=512^83/73
=1^83*8/73
=8 ans

Is this solution Helpfull? Yes (7) | No (8)  

Vivek kumar sahoo  (7 years ago)

(3^4)^250/73=(81^250)/73=(8^250)/73=((8^3)^83*8)/73=8

Is this solution Helpfull? Yes (2) | No (8)  

ayush  (7 years ago)

ans=8
3^1000/73=
from "fermet theorem" 3^72/73=1remainder; { x^(n-1)/n=1(remainder)}
so 3^(72*13)*3^64/73 3^72*13/73=1rem
= 3^64/73
= (3^4)*16/73 81/73=8rem
= 8^16/73
= (8^4)*4/73 8^4/73=8rem
= 8^4/73
gives remainder=8 ans

Is this solution Helpfull? Yes (2) | No (1)  

kadambala prudhvi  (7 years ago)

29 is the answer
3^1000/73=
=(3^4)250/73
=(81*250)/73
=((73+8)*250)/73
=250+(2000)/73
=250+(1460+540)/73
=250+20+540/73
=270+(73*7+29)/73
=277+29/73
so 29 is the answer

Is this solution Helpfull? Yes (1) | No (9)  

Akkili  (7 years ago)

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Kaviti Anjana  (7 years ago)

26 1000^3 and 1000 divided by 73 gives same remainder. So,1000 divided by 73 gives 26

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