Cookee Belen wrote:Sorry for posting in the wrong place (just started here). But thank you in advance for helping! Here's the problem: If n is a positive integer, is (1/10)^n <0.01?
(1) n>2
(2) (1/10)^n-1 < 0.1
I thought only A would be sufficient. But turns out either statement is.
Yeah, next time put it in the math section, but I'll answer it here.
Per my response to your general question on DS, put the given equation into a more useful form:
(1/10)^n = 10^-n
0.01 = 10^-2
So...
10^-n < 10^-2
Solving for n, the question is really asking: Is n > 2?
Knowing this, part 1 is obviously sufficient. Now look at part 2 and manipulate:
(1/10)^(n-1) < 10^-1
(1/10)^(n-1) = [(1/10)^n] * (1/10)^-1 = (10^-n) * 10
Dividing each side by 10, you get:
10^-n < 10^-2
This is a restatement of the second to last step in the original problem's manipulation above, so it answers the question and is sufficient.