Latest TCS Aptitude Question SOLUTION: 8. Find the last two digits of (1941^3881)*(1961^4181)

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(#M40011980) TCS question Keep an EYE

8. Find the last two digits of (1941^3881)*(1961^4181)

Asked In TCS Arit (11 years ago)

Unsolved Read Solution (9)

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MP   (11 years ago)

I think last two digits will be 01.

Is this solution Helpfull? Yes (19) | No (1)  

Amish Choudhary  (11 years ago)

taking last two digits of first term i.e. (41)^3881 now , taking (41)^1 = 41,
(41)^2=1681, (41)^3=68921, (41)^4=2825761, (41)^5=115856201.......so on... Do not write whole terms as i written take only last two digits we get a sequence of last digits-> 41,81,21,61,01, again 41,...... now by observation 81st(^3881) term is: 41, Similarly for second term(1961^4181) the sequence of last digits is-> 61,21,81,41,01, again 61,2...... now again by observation 61 is the 81st (^4181)now multiply 41 and 61 we get last two digits as 01.

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ajeet singh   (11 years ago)

( 41 ^81 ) * ( 61^81 )

= 41 * 61
= 01 last two digit

Is this solution Helpfull? Yes (8) | No   

Rajesh  (11 years ago)

According to my calculation i think the answer must be 01.

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

Atul  (11 years ago)

last digit is definitely 1.
so answer can not have options as 12,22,42,44.

I think Mp is right.
01 will be last two digits.

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anusha  (11 years ago)

let us take the last 2 digits in the given numbers 1941 i.e, 41 & 1961 i.e,61 so 41^3881=41 & 61^4181=61..now will multiply 41 & 61 so final solution for this is "01"

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Arit  (11 years ago)

but options are 12,22,42,44

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snehil  (11 years ago)

(1941^3881) gives 4*1=4 ie 41 at last digits
(1961^4181) gives 6*1=6 ie 61 as last digit
so total product will give 41*61= ..01 as last digits

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Manju  (11 years ago)

it is coming 01

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