I was wondering how many base cases are needed and when to stop (in general). For example, I have 4-cent and 11-cent stamps and I need to determine the amount of postage I can make, the cases I have that work are: $4,8,11,12,15,16,17,19,20,22,23,24,26,27,28,30 \geq$
Now what I'm stuck on is: what stops me from adding 31 to that pile?
On
The strong induction you might be thinking of for this problem would start with the four cases: \begin{align} 30&=11+11+4+4 \\ 31&=11+4+4+4+4+4 \\ 32&=4+4+4+4+4+4+4+4 \\ 33&=11+11+11 \\ \end{align}
Then any subsequent case $k>33$ can run induction on the basis that there is a solution for $k-4$ and add a 4-cent stamp to that solution.
On
Nothing at all is stopping you from "adding 31 to that pile." In fact, I believe your goal is to show that you can form $n$ cents of postage whenever $n\geq30$ (when you have just 4-cent and 11-cent stamps at your disposal).
Let $P(n)$ be the statement that you can form $n\geq30$ cents of postage using exclusively 4-cent and 11-cent stamps. Let $P(k)$ be the inductive hypothesis, the assumption that you can form $k\geq30$ cents of postage. Your goal, then, is to show that you can form $k+1$ cents of postage (that is, $P(k)\to P(k+1)$). You have already handled the base case so we will worry about the inductive argument.
As Hagen notes, if the $k$ cents had an 11-cent stamp, then you could easily replace it by three 4-cent stamps, forming $k+1$ cents. Otherwise, your $k$ cents of postage was formed using only 4-cent stamps. By the inductive hypothesis, with $k\geq30$, there must be at least eight 4-cent stamps in the mix. Now use three 11-cent stamps to replace eight 4-cent stamps--you have then formed $k+1$ cents.
Note that $P(n)$ is true for $n=30,31,32,33$ by our work above. Now assume $P(\ell)$ is true for all $\ell$ where $30\leq\ell\leq k$, where $k\geq33$. Since $k-3\geq30$, it must be the case that $P(k-3)$ is true (i.e., we may form $k-3$ cents of postage). Now simply add a 4-cent stamp and you have formed $k+1$ cents, as desired, thus proving $P(k+1)$ to be true as well. $\blacksquare$
If only $4$ and $11$ cent stamps are available and you know you can create $n$ cent, what would allow you to create $n+1$ sents? If the way you obtain $n$ cents involves at least one 11-cent stamp, you can replace it with three 4-cents. Or if it involves eight 4-cents, you can replace these with three 11-cents. But in general you do not know if either of these is the case. However, if $n\ge 32$ then you know that there actually either are at least eight 4-cents or there is at least one 11-cent (cause the 4-cents can only add up to $28$). Since we can make $32$ cents, this shows that all $n\ge 32$ are possible.
The smaller cases need to be considered separately. And this is readily done by simply trying zero up to at most two 11-cents and seeing if the rest is a multiple of $4$ (which works for $n=31$; in fact, since $30=11+11+4+4$, you obtain $31=11+4+4+4+4+4$ by the method described above).