How would I find a number where $\sum_{d\mid n}d > 100n$? How would I find an $n$ such that
$$
\sum_{d\mid n} d > 100 n
$$
I have absolutely no idea as to how to begin this question.
 A: You want lots of distinct prime divisors to have more divisors in total. A duplicate $p$ generally makes your $n$ larger without contributing as much to the sum as the next unused prime would. Try looking at $2\cdot3\cdot5\cdots p_k$.
In fact, if $n=2\cdot3\cdot5\cdots p_k$, then $$\sum_{d\mid n}d=\prod_{i=1}^k(p_i+1)$$ which we would like to be greater than $100\prod\limits_{i=1}^kp_i$. So we want $$\prod_{i=1}^k(1+1/p_i)>100$$ 
Now $$
\begin{align}
\prod_{i=1}^k(1+1/p_i)& >1+\sum_{i=1}^k\frac{1}{p_i}\\
\end{align}
$$ 
So it would suffice to have a value of $k$ large enough for the $k$th partial sum of the harmonic prime series to surpass $99$. This will eventually happen since the series diverges, but it does so even slower than the harmonic series. So it will take quite a while. Since $\sum\limits_{i=1}^k\frac{1}{p_i}$ is asymptotic to $\log(\log(k))$, $k$ will need to be in the neighborhood of $e^{e^{99}}$. But this is an overestimate with a lot lost when the product was replaced with the sum. You could get away with much smaller (although still enormous) $k$.
If you replace the $100$ by $10$, and if you trust WolframAlpha, then $k=553$ would work.

Added later: I believe that if $k$ is large enough to make the harmonic prime series's $k$th partial sum larger than 13.13 then you will get a ratio above 100.
As above, I am working with $n=2\cdot3\cdots p_k$. As above, we want $\prod_{i=1}^k(1+1/p_i)>100$. By expanding the product to its second-order terms, we can get this result.
Let $S_k=\frac12+\frac13+\cdots\frac{1}{p_k}$. Then $$S_k^2=\overbrace{\frac{1}{2^2}+\frac{1}{3^2}+\cdots\frac{1}{p_k^2}}^{T_k}+2\overbrace{\left(\frac1{2\cdot3}+\frac1{2\cdot5}+\cdots+\frac1{p_{k-1}\cdot p_k}\right)}^{U_k}$$
Note that $T_k$ is bounded above by $c=\zeta(2)-1=\frac{\pi^2}{6}-1\approx0.645$.
Now
$$
\begin{align}
\prod_{i=1}^k(1+1/p_i) & >1+S_k+U_k\\
& =1+S_k+\left(\frac{S_k^2-T_k}{2}\right)\\
& >1+S_k+\left(\frac{S_k^2-c}{2}\right)\\
\end{align}
$$
Applying the quadratic formula, this expression exceeds $100$ when $S_k>-1+\sqrt{199+c}\approx13.1295765707\ldots$. So if you can get $S_k$ this large, the corresponding $n$ should work. That roughly reduces $k$ to about $e^{e^{13.13}}$. There are obvious ways to lower this bound: $c$ can be shaved closer to the true sum of prime square reciprocals, or we could move on to a third-order expansion which would involve $\zeta(3)$ and bring the $13.13$ figure down to something closer to $\sqrt[3]{6\cdot100}$. I applied these and unless I made an algebra error during the third-order expansion, it brought $k$ down to $e^{e^{7.3}}$.
A: Claim: If $n = \prod p_i ^{a _i} $, then $\sum_{d \mid n} d = \prod \frac { p_i^{a_i+1} - 1 } { p_i - 1} $.
You should be aware of this from Number Theory.
Hence, if we want $\frac { \sum _{d \mid n} d} {n} > 100$, it is equivalent to $\prod\frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i} }> 100$.
Fact 1: As $a_i \rightarrow \infty$, then $\frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i}}  \rightarrow \frac { p_i}{(p_i - 1)}$.
Fact 2: 
$$ \prod_{p < n, p \mbox{ prime}} \frac {p}{p-1} \rightarrow \infty.$$
These two facts combined guarantee that such an $n$ exists. However, it will be very large.
