Smallest integer $n$ such that $\left(1-\frac{n}{365}\right)^n < \frac{1}{2}$ Find the smallest integer $n$ such that $$\left(1-\frac{n}{365}\right)^n < \frac{1}{2}.$$
I cannot use a calculator, and I do not know where to begin.
 A: To find an integer $n$ such that this holds, rewrite the inequality as $1-(n/365)<\exp(-(\ln2)/n)$ and use the fact that $\exp(-x)>1-x$ for every $x$. Then $n$ will do as soon as
$1-(n/365)<1-(\ln2/n)$, that is, $n^2>365\cdot\ln2$. Since $\ln2<.7$, one knows that $365\cdot\ln2<365\cdot.7=255.5<256=2^8$, hence every $n\geqslant2^4=16$ will do.
The numerical values the reasoning above requires to know to be performed without a calculator are the fact that $\ln2$ is (just) below $.7$ and the first powers of $2$.
It happens that the inequality does not hold for $n=15$ hence $n=16$ is the correct answer but at the moment I do not know how to prove this part without a calculator, except using the (alternating) expansion of the exponential at the second order to lower bound it. This is cumbersome, but here we go.
Since $\exp(-x)<1-x+\frac12x^2$ for every nonnegative $x$, it is enough to check that for $n=15$, $n^2<365\cdot\ln2\cdot(1-\ln2/n)$, which is true if $\ln2>\frac{n}2\left(1-\sqrt{1-\frac{4n}{365}}\right)$. Using $\sqrt{1+x}<1+\frac12x$ for $x=\frac{4n}{365-4n}$ yields $\sqrt{1-\frac{4n}{365}}>\frac{365-4n}{365-2n}$. Hence $\ln2>\frac{n^2}{365-2n}$ is enough, that is, for $n=15$, $\ln2>\frac{45}{67}$. Since $\frac{45}{67}\approx.672$, this proves the thing if one knows that $\ln2$ is greater than $.68$.
The numerical value the reasoning above requires to know to be performed without a calculator is the fact that $\ln2$ is about $.69$.
A: You could get an initial guess, by 
$$
\begin{aligned}
   &&\left(1-\frac{n}{365}\right)^n &< \frac{1}{2}.\\
   &\Leftrightarrow & n\ln\left(1-\frac{n}{365}\right) &< -\ln 2\\
\end{aligned}
$$
and using $\ln(1 + x)\approx x$ for small $x$ to arrive at $n\approx \sqrt{365\ln2}$. Then, you'd need to guess a value for this square root, and check if it really meets your requirements.
