Pie Pie Pie

Let $G_n$ be the partition of pie that $n^{th}$ person get, $F_n$ be the remaining Pie after $n^{th}$ person get their pie.

$F_n = \cfrac{100-n}{100} \cdot F_{n-1} \hspace{20mm} (1) \\ G_n = \cfrac{n}{100} \cdot F_{n-1} \hspace{26mm} (2)$ Rearrange (1),

$F_{n-1} = \cfrac{100}{100-n} \cdot F_{n} \hspace{20mm} (3)$

Compare $G_{n+1}$ and $G_{n}$ ,

$\begin{aligned} H(n) &= \cfrac{G_{n+1}}{G_n} = \cfrac{ \cfrac{n+1}{100} \cdot F_{n} } { \cfrac{n}{100} \cdot F_{n-1} } \hspace{10mm} \\ &=\cfrac{ \cfrac{n+1}{100} \cdot \cancel{F_{n}} } { \cfrac{n}{100-n} \cdot \cancel{ F_{n} } } \hspace{10mm} \text{sub (3)} \\ &=\cfrac{ (n+1)(100-n) }{ 100n } \\ &=\cfrac{ -n^2+99n+100 }{ 100n } \hspace{10mm} \text{H is monotonic decreasing as } \left( H'(n) = -\cfrac{1}{n^2} - \cfrac{1}{100} < 0 \right) \end{aligned}$

With $H(1) = \cfrac{198}{100} = 1.98$ and $H$ is monotonic decreasing, we check which $n$ will make $H(n) < 1$ ; before that $G_{n} < G_{n+1}$

$\begin{aligned} H(n) &=\cfrac{ -n^2+99n+100 }{ 100n } < 1 \\ &\implies -n^2+99n+100 < 100n \\ &\implies n^2+n-100 > 0 \\ &\implies n = \cfrac{-1+\sqrt{1+400}}{2} > \cfrac{-1+\sqrt{400}}{2} = \cfrac{-1+ 20}{2} = 9.5 \hspace{5mm} \text{neglect the -ve } \\ \end{aligned}$

It means when $n>9.5, H(n) = \cfrac{G_{n+1}}{G_{n}}< 1$ which means when $n\geq 10$ next person will get less pie than previous person.

Therefore, $H(9) = \cfrac{G_{10}}{G_{9}} > 1 \rightarrow G_{10} > G_{9} \therefore$ the $\boxed{10^{th} }$ person will get the most pie.

The problem is equivalent to finding integer $n$ for which $\dfrac {n(99)(98)(97) ... (101 - n)}{100^n}$ is maximum.

Can anyone tell me your approach of finding $n$ ?