Quantitative Aptitude :: Simplification - Discussion

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Explanation:
Last digit of 4^1 = 4 ; 4^3 = 4 ; 4^5= 4....
So, 99 = 1+(n-1)*2 n = 50 i.e. 4+4+4....50 times i.e. 4*50 = 200 (last digit 0)
Similarly, last digit of 9^2=1 ; 9^4 = 1....
So, 100 = 2+(n-1)*2 n=50 i.e. 9+9+9+....50 times i.e. 9*50 = 450 (last digit 0)
So, 200+450 = 650 i.e. last digit is '0'.