Latest TCS Aptitude Question SOLUTION: Last two digits of the expression (1141^3843) + (1961^4181) ?

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Last two digits of the expression (1141^3843) + (1961^4181) ?

Asked In TCS Nakshatra Mukhopadhyay (8 years ago)

Unsolved Read Solution (6)

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anoop mandal  (8 years ago)

Unit digit of this expression is always 1 as the base ends with 1. For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.
last two digits of (1141^3843)=(4*3),1=21
last two digits of (1961^4181)=(6*1),1=61
add 61+21
so ans = 82

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milan jena  (8 years ago)

1141^3843=21
1961^4181=61
21+61=82

Is this solution Helpfull? Yes (3) | No (1)  

Ramakrishna Siluveru  (8 years ago)

we should use power cycle method
take last 2 digit of 1141
41^1=41 r=1
41^2=81(take last 2 digits of ans) r=2
41^3=21 r=3
41^4=61 r=4
41^5=01 r=0
41^6=41
hence power cycle is repeating and it is 5
similarly for 1961
61^1=61 r=1
61^2=21 r=2
61^3=81 r=3
61^4=41 r=4
61^5=01 r=0
61^6=61
hence for 1961 powercycle is 5
for 1141 powercycle-5
4181mod5=rem1=61
for 1961 powercycle is 5
3843mod5=rem 3=21
therefore 21+61=82

Is this solution Helpfull? Yes (2) | No (3)  

ronaldo  (8 years ago)

unit digit is 1^3 so 1 and and tens digit is 4^3 take that unit digit so 2 like other side
ans is 82

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

radhika  (8 years ago)

Unit digit of this expression is always 1 as the base ends with 1. For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit
so ans = 82

Is this solution Helpfull? Yes | No (1)  

V.J.Sarath kumar  (8 years ago)

take last two digit of 1141 -> 4 x 3(unit digit of power) and (1^3 (unit digit of power)) = 21
same for 1961 u get 61. then add both u get 82.

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