When solving puzzles I often think the most frustrating thing is that there is very little theory about how to approach and efficiently solve them. Yet this is maybe why they are so interesting and allow one to be really creative. But sometimes I wish there had been more reasoning about the methodology of solving puzzles in my learning career than the sole perpetual advice that skill comes with practice.
Understand the question, see the challenge
I remember having spent a lot of time just figuring out what the heck this puzzle is asking for, in what direction the questions want me to go. This can of course be part of the puzzle itself, and maybe it is not that important after all to find the exact solution as fast as possible but rather just start thinking and go where you want to go. Only that degrees, school and partially also university are not always the right place for that.
The image in the mirror
A simple question I once heard and already asked some people is: "Why is left and right flipped in the mirror?" I then sometimes get a lesson in optics, how the light is reflected, a schematic drawing and the final answer that this is why of course left and right is mirrored, "You see?!". To be fair the question is not that obviously asking for a puzzle, but to make it more clear I could ask "Why of all things is it left and right? Why not top and bottom?" Then the puzzling question may become more obvious. I started to think about what two images are actually compared, that it is easier to not take myself as the subject to think with but somebody else I am looking at, once directly standing in front and once in the mirror. Then I thought about how I come from one image to the other and what needs to be changed to make top and bottom flip instead.
I found this example particularly interesting because at first it seems to ask for a simple technical explanation, but then when you see the actual implied challenge it gets much more puzzling.
No need to solve what is not asked for
Maybe not an advice I would give for trying to be creative, but unarguably efficiency is sometimes what is asked for.
One glass of water and one glass of orange juice
Assume both glasses a filled with the same amount. If you take one spoonful of the water glass, put it into the orange juice, mix it, take a spoonful of the mix and put it back into the water, will there be more orange juice in the water or more water in the orange juice or the same amount of each one in the other (disregarding density of the different fluids)? When I first started to solve this I made the mistake to actually starting to compute the fractions of each fluid in the other. While you may be able to solve it that way, remember that this is more than is asked for. The question is only if the parts are equal or in case not, which one is bigger.
There is actually a high level thought to easily solve this. If you take one spoonful from one glass to the other and then a spoonful of the same size again back, there must be the same total amount in each glass as before. Knowing that, if amount x is missing in the water glass because it is in the orange juice, the same amount of juice must be in the water glass, otherwise they would not have the same equal amount as they had before. Of course it can also be interesting to compute the actual fractions but it is a less efficient way to solve the original question.
Ask the neighbors
Sometimes a seeming detour can be a shortcut. It may be easier to compute something else, perhaps similar, and go on from there.
Power set
What is the sum of 2^0 + 2^1 + 2^2 + ... + 2^k? Not trivial to convert but in another representation it looks much simpler. This is of course the representation of a binary number with k digits each being 1. E.g. for k=4 it is 1111. To bring it into a simpler form the lots of 1 are bothering. Can this number be represented as a sum that has much less 1, maybe a fixed amount of 1, not depending on k? Well, 1111 is the same as 10000-1. And this form always has only two 1. So sum[i=0, k, 2^i] = 2^(k+1)-1.
The mathematical conversion for the interesting part would go something like this: 2^(k+1) = 2 * 2^k = 2^k + 2^k = 2^k + 2 * 2^(k-1) = 2^k + 2^(k-1) + 2^(k-1) = 2^k + 2^(k-1) + 2 * 2^(k-2) = ....