maipenrai wrote:
Ok lets try this one.
\(\ 3^{123}= \ (5-2)^{123}= \ 5^{123}+(-2)^{123}\)
\(\ 5^{123}\) is always divisible by 5.
So need to look at the \(\ -2^{123}\)
Units digit of -2 to the power of n is cyclical and consists of 4 possibilities (-2, 4, -8, 6)
Hence \(\ (-2)^{120}\) will have last digit 6 and \(\ (-2)^{123}\) = -8.
Dividing 8 by -5 we have remainder
8=-5n+R
For n=2
8=-5*2+R
R=2
Hence Answer is (C)
hi,
another approach after the highlighted portion
So need to look at the \(\ -2^{123}\)
\(\ -2^{123}= (-2^4)^{\frac{123}{4}} = 16^{30}*-2^3=(15+1)^{30}*-8\)
now\((15+1)^{30}\) will leave 1 as remainder, so total remainder = 1*-8 = -8..
as correctly done, we cannot have negative remainder so remainder = 10-8 = 2
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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77% (00:44) correct 
