nsuen wrote:HI-

I could not understand the explanation of OG 2015 DS 41, in particular, i dont understand the part that said..greater than 10^n if 1 is greater than 10^2n, I tried to test case using mahattan prep method but couldnt get the right answer.



Thanks for your help in advance.

The question: If n is an integer is (.1)^n greater than (10)^n?

Let's rephrase. .1 = (10)^-1. So (.1)^n = [(10)^-1]^n = (10)^-n. Rephrased question Is (10)^-n > (10)^n?

S1: n > -10.
Case 1: n = -9. (10)^-n = (10)^9 and (10)^n = (10)^-9. (10)^9 is greater than (10)^-9 so the answer to Is (10)^-n > (10)^n? is YES.
Case 2: n = 9. (10)^-n = (10)^-9 and (10)^n = (10)^9. (10)^-9 is NOT greater than (10)^9 so the answer to Is (10)^-n > (10)^n? is NO.
Because we can get a YES and a NO, this statement is not sufficient.

S2: n< 10
We can reuse both cases that we used in statement 1, because 9 and -9 are both less than 10, so we already know we can get a YES and a NO. So this statement alone is not sufficient, and the statements together are not sufficient. Answer is E

Dear David and Rich,

Thank you very much, I understand now.

Rich: how do you just know to use 0 and -1 will be the quickest way? Is it something I will gain better as practice or is there a strategy that makes you pick 0 and -1?

nsuen wrote:Dear David and Rich,

Thank you very much, I understand now.

Rich: how do you just know to use 0 and -1 will be the quickest way? Is it something I will gain better as practice or is there a strategy that makes you pick 0 and -1?

Generally speaking, you want to pick numbers that yield simple calculations. 0 is always easy. And if we see that one term is raised to a negative exponent, and the other is not, we'll want to pick a negative number too, as this will be likely to yield a different result. -1 is easiest.