What is the last digit of 9^1+99^2+999^3+...+(10^n-1)^n
We have the sum of integers:
(..9) + (..1) + (..9) + (..1) + ...
Statement (1) by itself is sufficient. If n is even then this sum ends with 0. If it's odd, then this sum ends with 9.
Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd)
it seems from question's stem that n>3, which combined with Statement (2) means that n is odd. As a result, the sum always ends with 9. If this correct, the answer should be D.
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