Latest TCS Aptitude Question SOLUTION: find last 3 digits.......2988^687^241?

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(#M40026218) TCS Logical Reasoning question Keep an EYE

find last 3 digits.......2988^687^241?

Asked In TCS MD (10 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Prateek Pandey  (10 years ago)

After reading this, you'll kill yourself
2988^687^241 mod 1000
=988^687^241 mod 1000
let 687^241 be n
988^n mod 1000
247^n mod 1000 * 4^n mod 1000
first for 247^n mod 1000
lambda(1000)=100
so, we need 687^241 mod 100
lambda(100)=20
241 mod 20=1
687^1 mod 100 = 87 mod 100 or -13 mod 100
now,
247^(-13) mod 1000
multiplicative inv of 247 is 583
583^13 mod 1000
863
now for 4^n mod 1000
=(2^(2n-3) mod 125)*8
lambda(125)=100
So, we need 687^241 mod 100 = 87 mod 100 (already calculated)
So n=87
2*n-3= 171
(2^171 mod 125)*8
=(2^71 mod 125)*8 (Since 171 mod 100 =71)
reduce using 2^(10*7+1) mod 125
you'll get
(98 mod 125)*8
=784
now 784*863 mod 1000
=592(ans)

Is this solution Helpfull? Yes (4) | No (16)  

MOUMI  (10 years ago)

512 is the answer.

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

lavanya  (10 years ago)

Just explain in simple way

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

sourav mukherjee  (10 years ago)

272 is d ans...

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

jitendra kumar  (10 years ago)

@prateek bhai jo tm concept diye ho kafi kara h

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