Cross a Magical Bridge Infinite Times

Suppose that the toll is $t$ coins and that you start with $x$ coins.

The first time you cross the bridge, you have $x-t>0$ coins left.

The second time, you have $2(x-t)-t>0$ coins left.

The third time, you have $3(2(x-t)-t)-t>0$ coins left.

The $n$ th time, you have $n((n-1)((n-2)(\cdots 3(2(x-t)-t)-t)-t\cdots)-t)-t>0$ coins left. Simplify this to get $n!\ x-n!\ t-\frac{n!}{2!}t-\frac{n!}{3!}t-\cdots-\frac{n!}{n!}t>0$ . Then, solve it to get that $x>t\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots\frac{1}{n!}\right)$ for all $n$ . Note that this is a bounded, monotonically increasing sequence.

When $n\to\infty$ , we find that $x>t(e-1)$ .

The minimum positive integer $x$ when $t=10000$ and $n\to\infty$ is therefore $17183$ coins.

Javascript:

for (let init = 0;; init++) {

let coin = init

for (let n = 1; n < 100; n++) {

coin = coin * n - 10000

if (coin < 0) {

break

}

}

if (coin > 0) {

console.log(init)

break

}

}

I also have a solution use Gamma function...