How to reach 1000 reputation? I wonder how many ways there are to reach 1000 reputation on Math SE in $N$ steps.
Each way is a unique combination of the following additions or deductions, e.g. $+10$ 89 times along with $(+5, +2,+2)$. This way has 92 steps.
One step is an addition or a deduction.
For additions, there are
$$(+1, +2, +5, +10, +15)$$
For deductions, there are
$$(-1, -2, -5, -15, -100)$$
With 1 base point and 100 points for trusting, this is equivalent to asking how many positive integer solutions there are for
$$101+(a_1+2a_2+5a_3+10a_4+15a_5)-(b_1+2b_2+5b_3+15b_4+100b_5)=1000$$
with
$$\sum a_n +\sum b_n\le N$$
Moreover, if $s(N)$ is the solution-counting function, what is the asymptotic behaviour for large $N$?
I have no idea of how this problem can be solved.
Any suggestions?
Thanks in advance.
 A: This is an attempt.  I am very well aware that you cannot get arbitrarily large negative total reputations, but let us assume for the sake of simplicity that this can happen.  However, I have a strong hope that the correct asymptotic form (with the lower bound $1$ is taken into account) should be close to what I have obtained.
As for the asymptotic behavior, one can use a probabilistic argument as follows.  Let $R\subseteq \mathbb{Z}_{>0}$ be the set of all possible absolute reputation points one can gain or lose.  Suppose that $m:=|R|$.  Let $\mathbb{P}$ be the discrete uniform probability measure on $\Omega:=R\cup (-R)$.  
Let $X_n$ be the random variable of the reputation change at step $n\in\mathbb{Z}_{>0}$ ($X_n$ takes value in $\Omega$, and is negative for reputation losses).  Assume that the random variables $X_n$'s are independent and identically distributed with the uniform discrete distribution on $\Omega$.  Thus, the expected value of each $X_n$ is $\mathbb{E}[X_n]=0$, whereas the standard deviation $\text{stdev}(X_n)$ is
$$\sigma:=\sqrt{\frac{1}{m}\,\sum_{r\in R}\,r^2}\,.$$
Write $S_n:=X_1+X_2+\ldots+X_n$ for every $n=1,2,3,\ldots$.  By the Central Limit Theorem (CLT), the random variables $Y_n:=\dfrac{S_n}{\sigma\,\sqrt{n}}$ for $n\in\mathbb{Z}_{>0}$ converge in distribution to the standard normal variable.  That is, for each $x\in \mathbb{R}$,
$$\lim_{n\to\infty}\,\mathbb{P}[Y_n\leq x]=\frac{1}{2}\,\Biggl(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Biggr)\,.$$
Hence, 
$$\lim_{n\to\infty}\,\mathbb{P}\left[S_n\leq \sigma\,\sqrt{n}\,x\right]=
\frac{1}{2}\,\Biggl(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Biggr)\,.$$
Thus, we get
$$\mathbb{P}\left[\sigma\,\sqrt{n}\,(x-\epsilon)<S_n\leq \sigma\,\sqrt{n}\,x\right]\approx \frac{1}{\sqrt{2\pi}}\,\exp\left(-\frac{x^2}{2}\right)\,\epsilon\,,\tag{*}$$
where $\epsilon>0$ is small and $n$ is a large positive integer.  
Now, for a target score $t\in\mathbb{Z}$, we have by setting $x:=\dfrac{t}{\sigma\,\sqrt{n}}$ and $\epsilon:=\dfrac{1}{\sigma\,\sqrt{n}}$ in (*) that
$$\mathbb{P}\left[t-1<S_n\leq t\right]\approx \frac{1}{\sqrt{2\pi\,\sigma^2\,n}}\,\exp\left(-\frac{t^2}{2n\sigma^2}\right)\,.$$
In other words, the number of ways to get $S_n=t$ for a fixed integer $t$ and for a large integer $n>0$ is
$$f(n,t):=|\Omega|^n\,\mathbb{P}\left[t-1<S_n\leq t\right]\approx \frac{(2m)^n}{\sqrt{2\pi\,\sigma^2\,n}}\,\exp\left(-\frac{t^2}{2n\sigma^2}\right)\,.$$
If $|t|\ll \sigma\,\sqrt{n}$, then we may further say that
$$f(n,t)\approx \frac{(2m)^n}{\sqrt{2\pi\,\sigma^2\,n}}\in \Theta\left(\frac{(2m)^n}{\sigma\,\sqrt{n}}\right)\,,$$
which is independent of $t$ (well, in the approximating sense).
Note that I also assume that the starting point is $0$.  If you start from $1$ and your final score is $1000$, then you should set $t:=1000-1$.  If you start from $101$ with the same final score $1000$, then set $t:=1000-101$.
