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(#M40029785) TCS Numerical Ability question Keep an EYE

Find last two digits of (1021^3921)+ (3081^3921)

Asked In TCS soniya shrivastava (11 years ago)

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RAKESH  (11 years ago)

unit place of both the no.s is 1
10th place of 1021^3921=10th place of 1021*unit place of 3921=2*1=2
10th place of 3081^3921=10th place of 3081*unit place of 3921=8*1=8
so last two digits of (1021^3921)=21
and last two digit of (3081^3921)=81
so 21+81=102
hence last two digits of (1021^3921)+ (3081^3921)=02

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Shuvadip Lahiri  (11 years ago)

1021^3921=21^3921=21
3081^3921=81^3921=81
last two digit=81+21=02..

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Gaurav Singhal   (11 years ago)

3081=980*4+1 so according to cyclicity 21+81=102..ans is 02

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Smart  (11 years ago)

period if 21 and 81 is 5
hence 21^3921=21 and 81^3921=81
81+21=01
hence 01 is answer

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Dhanashree Shirke  (11 years ago)

(3921/4)remainder=1
so 1021^1+3081^1=4102
so last two digit = 02

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Guru Pranav  (11 years ago)

02

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