Why/when is more than one base case needed in mathematical induction? I've recently started studying mathematical induction and came across this proof:Could someone please explain to me why 10 base cases are necessary? When is more than one base case necessary?
 A: I disagree with the claim that there are ten base cases here. The base case is $n=1$, which is just one single case! It just so happens that this case can be proved by splitting it into ten cases. Though this isn't necessary: the sum of all digits of the number $mmm$, $m\in\{0,1,\dots,9\}$ is $m+m+m=3m$, which is divisible by $3$, so the number itself is also divisible by $3$. No casework needed.
But your question is more general. As a rule of thumb, induction can be used when the sequence of objects we're considering can be written recursively. For instance, the sum $s_n:=\sum_{k=0}^nk$ can be written recursively as $s_{n+1}=s_{n}+n+1$, and then we can use induction to prove the formula $s_n=\frac{n(n+1)}{2}$: The base case $n=0$ is true, and then we have as the inductive step
$$s_{n+1}=s_n+n+1=\frac{n(n+1)}{2}+n+1=\dots=\frac{(n+1)(n+2)}{2},$$
where the crucial step for induction was in expressing our object of interest in a recursive fashion.
Now the number of base cases depends on our recursion formula. The formula above only references one prior element in the sequence. In this case, we only have to prove one base case. But consider this different setting: Let $t_n$ be the "Tribonacci" sequence (which I just made up) defined through $t_0=t_1=t_2=1$ and $t_n=t_{n-1}+t_{n-2}+t_{n-3}$ for all other members of the sequence. Here we reference three prior elements of the sequence, so to use induction to prove facts about this sequence we need three base cases. For instance, we can prove that all Tribonacci numbers are odd. Proof: It's true for the base cases $t_0,t_1,t_2$ by definition. Now the inductive step uses the oddness of all three prior elements: $t_n$ is the sum of three integers. If all three are odd, then their sum is odd. This hinges on the fact that all three summands are odd, so we have to establish the oddness of all three initial summands in the base case, which means we have to cover three elements of the sequence before the induction carries over to all other elements of the sequence.
A: I think this is a nice question for those new to writing proofs, as induction is quite an important proof technique. I'll note a few things here.
First, it is important to note what you are inducting on. In the proof you linked to (as in the comments, I encourage you to use mathjax to type up the proof yourself), you are inducting on $n$ where $3n$ is the number of digits. So, the proof really only has one base case: the case where $n=1$. It is just that the base case of $n=1$ can occur in $10$ different ways, and the proof you linked to checks each such instance one at a time. Here is a proof of the $n=1$ base case that is a bit shorter.

Suppose $n=1$, meaning $kkk$ is a $3$-digit number with identical digits all being $k$. Then, we have $$kkk =  k \cdot 100 + k \cdot 10 + k = k(100+10+1) = k \cdot 111 = k \cdot 3 \cdot 37.$$ Therefore, $3$ divides $kkk$ as $3 \cdot (37 \cdot k) = kkk$.

This proof maybe helps solidify the idea that there really is only one base case: the case where $n=1$. We just handled it by using the variable $k$ instead of considering $000, 111, \ldots, 999$ each separately. It is very, very common in induction proofs to use variables as often times the base case(s) have infinitely many ways that they can occur (you may first run into such an induction proof in a linear algebra class, when inducting on the dimension of a vector subspace since often there are infinitely many subspaces that have the same dimension).
With all that being said, there are times where genuinely more than one base case is needed. This means that if you are inducting on some quantity $n \ge 1$, then you may have to check $n=1$ and $n =2$ and maybe even more cases. The number of base cases you need to check depends entirely on what the induction hypothesis and induction step need. For example, sometimes the induction hypothesis is of the form "assume something holds true for $n+1$ and $n$ where $n \ge 0$" and then you prove it for $n+2$. In this situation, you would need to prove the base cases of $n = 0$ and $n=1$ since neither $n=0$ or $n=1$ can follow from the induction step and both would be needed to prove $n=2$ via the inductive step.
I hope this helps, best of luck with your studies!

Addendum 1: Unfortunately, I cannot think of a simple example off the top of my head for an induction proof that actually requires two base cases. I am a bit pressed for time, so I'd like to especially encourage others (as always) to comment their examples below or write other answers with examples. Also please feel free to suggest how I can improve the above explanation.


Addendum 2: This question may also be a good opportunity to explain strong induction, as it is related or at least a strong induction proof may provide a simple example where multiple base cases would be needed

