Question: A birthday pie is divided among 8 guests, as follows. The first guest gets $1/8$th of the pie, the second guest $2/8$th of the remaining pie, etc. Who gets the biggest piece?
Can you solve it as well for 100 guests, where the $k$-th guest receives $k/100$ of the remaining pie?
Answer: The third guest gets the biggest piece. The shares received are:
1/8, 7/32, 63/256, 105/512, and smaller fractions for the later guests. 63/256 $\approx$ 25%, the largest piece of the pie.
In general the puzzle can formulated as follows. There are $N$ guests, and the $k$-th guest receives $k/N$ of the remaining pie. Now, let $x$ be the remaining fraction of the pie when the $k$-th guest is being served. The $k$-th guest thus receives $x \cdot \frac{k}{N}$, after which there is $x \cdot \frac{N-k}{N}$ left for guest $k+1$. Therefore the $k+1$-th guest receives $[x \cdot \frac{N-k}{N}] \cdot \frac{k+1}{N}$. Guest $k+1$ receives a larger fraction then guest $k$ if $[x \cdot \frac{N-k}{N}] \cdot \frac{k+1}{N} > x \cdot \frac{k}{N}$ which rearranges to
$-k^2 -k + N > 0$
For N=100 this is true for $k < 10$, which means the 10th guest receives the largest piece of the pie.
Source: Pythagoras Magazine