Carcass wrote:
What is the least integer \(k\) such that \(\frac{1}{3^k}\) is less than \(0.0005\) ?
A. 6
B. 7
C. 666
D. 667
E. There is no such least integers
First, we can rewrite \(0.0005\) as:
\(0.0005 = \frac{5}{10,000}\) = \(\frac{1}{2000}\)
This question boils down to: for what value of \(k\) will \(3^k > 2000\)?
We know this is possible, so E is incorrect. C and D would work, but we want the least number. We don't need 600+ multiples of 3 to get above 2000, so we can cross those out. We're left with A and B.
\(k = 2 => \frac{1}{9}\)
\(k = 3 => \frac{1}{27}\)
\(k = 4 => \frac{1}{81}\)
\(k = 5 => \frac{1}{243}\)
At this point, instead of continuing I'd estimate \(\frac{1}{243}\) to \(\frac{1}{250}\) in order to save time with calculation. Though by all means, if you have time to use a calculator or want to write out the multiplication, that's fine too.
\(k = 5 => ~\frac{1}{250}\)
\(k = 6 => ~\frac{1}{750}\)
\(k = 7 => ~\frac{1}{2250}\)
k = 7 is the least number for \(3^k > 2000\), so answer B is correct.