Latest TCS Aptitude Question SOLUTION: find out last two digits of (2957^3661)+(3081^3643) Options o 42 o 38 o 98 o 22

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find out last two digits of (2957^3661)+(3081^3643)
Options
o 42
o 38
o 98
o 22

Asked In TCS Srivastav Reddy (11 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

shyamlesh dash  (11 years ago)

haha i got the answer its 98 see i read this link http://www.totalgadha.com/mod/forum/discuss.php?d=4405
its like for the first number -------
57^3661=(57^4)^915*(57)=(49*49)^915*(57)=(2401)^915*9(57)
take tens digit which is 01 and then according to the trick from the link
=10*unit digit of power=01^anuthing is same and 01*57=57
second number 81^3643=8*3=24 so it is 41 and so adding 41+57=98

Is this solution Helpfull? Yes (21) | No (3)  

nagraj  (11 years ago)

yeah shyamlesh.... sorry every 1 logic z same but a mistake in calc...

2957)^1=...57 as last two digts
(2957)^2=...49
(2957)^3=...93
(2957)^4=...01
(2957)^5=...57


as we observed dat last digit repeats for every power of five...
now consider (2957)^3661=(2957)^{3660) * 2957
since 3660 is multiple of 4 it gives 01 as last two digits and again multiplying with 2957 we get 57 as last digits..

similarly do for (3081)^3643... we get 41 as last two dits

now finally adding both 57+41= 98 therefore 98 is last two digits...

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

nagraj  (11 years ago)

answer is 38..
observe carefully..
(2957)^1=...57 as last two digts
(2957)^2=...49
(2957)^3=...93
(2957)^4=...61
(2957)^5=...77
(2957)^6=...49

as we observed dat last digit repeats for every power of five...
now consider (2957)^3661=(2957)^{3660) * 2957
since 3660 is multiple of 4 it gives 61 as last two digits and again multiplying with 2957 we get 77 as last digits..

similarly do for (3081)^3643... we get 61 as last two dits

now finally adding both 77+61= 138 therefore 38 is last two digits....

Is this solution Helpfull? Yes (9) | No (13)  

shyamlesh dash  (11 years ago)

see it can also be solved as mentioned by nagraj for the frst term it is
57^1=57 ^2=49 ^3=93 ^4=01 and ^5=57 so it follows a cycle of 4 so 57*915+1=57 and so for second term is 47 which sums up to 98 :DDD

Is this solution Helpfull? Yes (4) | No (1)  

teja  (11 years ago)

sorry its 98 wrong calculation

Is this solution Helpfull? Yes (2) | No   

teja  (11 years ago)

ans is 42 i am sure about it

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