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n is a positive integer greater than 2. If y = 9^0 + 9^1+

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Source: — Data Sufficiency |

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by Ian Stewart » Sat Apr 06, 2019 1:18 pm

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If you look at the units digits of various powers of 9, you'll see they alternate between 1 and 9:

9^0 ends in 1
9^1 ends in 9
9^2 ends in 1
9^3 ends in 9

and so on. So if we add 9^0 + 9^1 + 9^2 + ... + 9^n, we'll get something ending in 0 when n is even, and something ending in 1 when n is odd. So Statement 2 is sufficient here (since if a number ends in 1, its remainder is 1 when we divide it by 5), while Statement 1 is irrelevant.
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