Christmas Streak 46/88: Nerds, Money, and Charity

Let's say there're $n$ nerds doing this.

And let's assume that the last nerd took $\$ k.$

Then before that nerd took it, there was $\$(nk+1).$

Then before the second last nerd took it, there was $\$\left(\dfrac{n}{n-1}(nk+1)+1\right)=\$\left(\dfrac{n^2k}{n-1}+\dfrac{n}{n-1}+1\right).$

Before the third last nerd took it, there was $\$\left(\dfrac{n}{n-1}\left(\dfrac{n^2k}{n-1}+\dfrac{n}{n-1}+1\right)+1\right)=\$\left(\dfrac{n^3k}{(n-1)^2}+\left(\dfrac{n}{n-1}\right)^2+\dfrac{n}{n-1}+1\right).$

So, before the first nerd took it, there was $\$\left(\dfrac{n^nk}{(n-1)^{n-1}}+\left(\dfrac{n}{n-1}\right)^{n-1}+~\cdots~+\dfrac{n}{n-1}+1\right)=\$\dfrac{n^nk+n^n-(n-1)^n}{(n-1)^{n-1}}.$

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Since the numerator must be divisible by the denominator, $n^nk+n^n-(n-1)^n\equiv 0\pmod{(n-1)^{n-1}}.$

Then $n^n(k+1)\equiv0\pmod{(n-1)^{n-1}}~\Leftrightarrow~k\equiv -1 \pmod{(n-1)^{n-1}}~\Leftrightarrow~k=(n-1)^{n-1}t-1.~(t\in\mathbb{N})$

But since the leftover money after the last nerd, $(n-1)k,$ should be divisible by $n,$ we get $k\equiv0\pmod{n}.$

$k=(n-1)^{n-1}t-1\equiv(-1)^{n-1}t-1\equiv0\pmod{n}~\Leftrightarrow~\cases{\begin{aligned} &k=(n-1)^{n-1}-1 &(n~\text{is odd})\\\\ &k=(n-1)^n-1 &(n~\text{is even})\end{aligned}}$

Therefore, the minimum amount of money the nerds earned in the first place is

$\cases{\begin{aligned} &n^n-n+1&(n~\text{is odd}) \\\\ &n^n(n-1)-n+1&(n~\text{is even})\end{aligned}}$

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Since $n=16$ in this problem, we obtain $\$(15\cdot 16^{16}-15)=\$(15\cdot 2^{64}-15).$

$\therefore~a+b=15+64=\boxed{79}.$