In mathematics there are often two ways of answering a question,
Lots of thinking and little working,
Little thinking and lots of working.
I was wondering what the most extreme cases of this would be? Questions where the "usual" answer is just to plough through the working and hope everything turns out all right, but a much more subtle solution exists (and ideally this subtle solution would point out what makes the question "tick", as it were).
For example,
Question: Prove that the symmetric difference operation is associative.
The symmetric difference of two sets, $S \triangle T$, is defined to be $(S\cup T)\setminus (S \cap T)$. Now, one could simply plough through the working, but the working is tedious and there is a "trick" half way through it. However, a much more subtle solution exists. It requires a bit of setting-up, but is overall much neater.
Proof: The idea is that $S\triangle T$ "looks like" addition mod $2$. To see this, notice that,
If $x\in S$, $x\in T$ then $x\not\in S\triangle T$, which corresponds to $1+1=0 \text{ mod }2$
If $x\in S$, $x\not\in T$ then $x\in S\triangle T$, which corresponds to $1+0=1\text{ mod }2$
If $x\not\in S$, $x\in T$ then $x\in S\triangle T$, which corresponds to $0+1=1\text{ mod }2$
If $x\not\in S$, $x\not\in T$ then $x\not\in S\triangle T$, which corresponds to $0+0=0\text{ mod }2$
This means that showing $(S\triangle T)\triangle U=S\triangle (T\triangle U)$ now comes down to showing that addition mod $2$ is associative. And it is, which is easily checked. So we're done.