Find the Largest Number from 50^99 and 99!
Which is the largest number from 50^99 and 99!
Where 50^99 = 50 × 50 × 50 ×……..× 50 {99 terms}
and 99! = 99 × 98 × 97 ×……..× 1
Solution: Finding the largest number
Let start with 50^99/99!
\begin{aligned}
\frac{50^{99}}{99!}&= \frac{50\times50\times50\times......\times50}{99\times98\times97\times......\times1}\\
\end{aligned}Rearrange the right-handed side
\begin{aligned}
\frac{50^{99}}{99!}&= \frac{50\times50}{99\times1}\times\frac{50\times50}{98\times2}\times\frac{50\times50}{97\times2}\times........\times\frac{50\times50}{51\times49}\times\frac{50}{50}\\
\\
&= \frac{50^2}{(50+49)(50-49)}\times\frac{50^2}{(50+48)(50-48)}\times........\times\frac{50^2}{(50+1)(50-1)}\times\frac{1}{1}\\
\end{aligned}We know (a + b)(a – b) = a² – b²
so
\begin{aligned}
\frac{50^{99}}{99!}&= \frac{50^2}{50^2-49^2}\times\frac{50^2}{50^2-48^2}\times........\times\frac{50^2}{50^2-1^2}\\
\end{aligned}Here
\begin{aligned}
\frac{50^2}{50^2-49^2}&>1 \\
\\
\frac{50^2}{50^2-48^2}&>1\\
&.\\
&.\\
&.\\
&.\\
&.\\
\frac{50^2}{50^2-1^2}&>1\\
\end{aligned}So we can write
\begin{aligned}
\frac{50^{99}}{99!}&>1\times1\times........\times1\\
\\
\frac{50^{99}}{99!}&>1\\
\\
50^{99}&>99!
\end{aligned}
