Latest TCS Aptitude Question SOLUTION: If X^Y denotes X raised to the power Y, Find the last two digits of ( 1941 ^ 3843 ) + ( 1961 ^ 4181). (a)12 (b) 22 (c) 42 (d)82

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(#M40024330) TCS Logical Reasoning question Keep an EYE

If X^Y denotes X raised to the power Y, Find the last two digits of ( 1941 ^ 3843 ) + ( 1961 ^ 4181).
(a)12
(b) 22
(c) 42
(d)82

Asked In TCS vulli tejaswi (11 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Lakshmi  (11 years ago)

last two digits of 41^1=41,41^2= 81,41^3=21,41^4=61,41^5=01,41^6=41,41^7=81.....
the same sequence repeats from 41^6
1941^3843=1941^5(768)+3=1941^3 gives 21 as last 2 digits.
By doing same procedure for 1961^4181 gives 61 as last 2 digits
then by summing them v get 21+61=82
ANS is 82

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Manohar Kumar  (11 years ago)

last term of 1941^3843 will be 1 and second last term will be 4*3=x2.
So last two term of the expression will be 21
Similarly the last two term of 1961^4181 will be 61.
So the last two term of the full expression will be 21+61=82.

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Animesh  (11 years ago)

ans will be 82

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Aamani Bonthala  (11 years ago)

ans : 82

This can be achieved by using power cycles concept. Power cycle of 41 is 5 and power cycle of 61 is also 5.Then in such case we need to divide the power of 41 by 5 and also divide power of 61 by 5.as they asked for last two digits we are considering only last two digits.Now if we divide we the remainders in both case as 3 and 1 respectively.Hence 41^3 leaves 21 in last two digits and 61^1 gives 61.Now add both 21 and 61 we get 82 as result.

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Gaurav Kumar  (11 years ago)

plz explain the process

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kajal jaiswal  (11 years ago)

I think its 22..since power cycle of both is 6 not 5.

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ponnada sowjanya  (11 years ago)

ans will be 82

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Sreelakshmi Gopinath  (5 years ago)

By taking the number 1941 ,
Last digit is 41 .so
41^1=41(ie, from the table given below)
If the last digit is 0 then put 1 in unit place
Similarily,
0- 1
1-1
2-4
3-4
4-2
5-1
6-1
7-4
8-4 ( byhearting this values become quite easy to solve)
So here 41^1=41
41^2=81
41^3=21
41^4=61
41^5=01 ( after that cycle repeats)
So we can write (1941^5(768)+3))
Here 768 is obtained by dividing 3843 by 5 & 3 is the remainder)
So 41^3 = 21
So 21 .
Then doing similar procedure in case of 1961 we get ans 61
So 21+61= 82


82 ans

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