Latest TCS Aptitude Question SOLUTION: Find last three digits of 2988^687

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Find last three digits of 2988^687

Asked In TCS Amrutha Duggirala (11 years ago)

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Y.Prasanna Chandra Lekha  (11 years ago)

(2988^687)/1000 leaves remainder as the last 3 digits of result
this can be = 988^(4*171+3)=988^3 by cyclicity
this leaves 272 as last 3 digits

Is this solution Helpfull? Yes (5) | No (14)  

tejomay saha  (11 years ago)

please give a proper solution

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

Durga  (11 years ago)

unit digit of 2988 is 8
The cyclicity of 2,3,7,8 is 4
The cyclicity of 0,1,5,6 is the no. itself
The cyclicity of 4,9 is 2
To check whether the number is divisible by 4 we have to divide the last two digit by 4. Likewise in 687 we have to divide 87 by 4 we get the remainder 3.So we have to put 3 as a power of unit digit i.e. 8(in base number)
8^3=432 therefore the ans is 432

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Lakshmi  (11 years ago)

@chandra lekha
Cycle is not repeating at power of 4 right?
then how u taken as multiple of 4 and all...
pls explain...

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ipsita saha  (11 years ago)

pls explain properly

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sahithi  (11 years ago)

@durga 8^3=512 not 432
so,ans:512

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Guru Pranav  (11 years ago)

mine came 592 :p

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Surojit Chinya  (11 years ago)

(2988^687)/1000 leaves remainder as the last 3 digits of result
this can be = 988^(4*171+3)=988^3 by cyclicity
this leaves 272 as last 3 digits

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Saumashankar Bala  (8 years ago)

ans is 504

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