algebra precalculus - What method for mentally computing 2-digit multiplication problems, minimizes the amount of mental steps?

So I've been practicing alot of mental math recently and ofcourse as a part of that, multiplying a double-digit number by another double-digit number. I have been doing some research into what the quickest way to computing the outcome of such a multiplication is but I still find myself having to go through too many steps in my head for each exercise. The goal is to have someone tell you a problem and then, without having them repeat the problem, computing the answer.

Let's take as an example: $63*88$

I will go over multiple methods that I know for computing this mentally, and explain my problems with each of them.

1) The elementary method. I think this is the first method of multiplying that everyone learns. It finds the answer by brute force multiplication and addition. Using this method on our example exercise would mean we take the following steps:

• First simplify the left factor and multiply it with the complete right factor, yielding:

$60*88=60*80+60*8=4800+480=5280$

• Then multiply the remaining $3$ with our right factor and add it to the result above, yielding:

$3*88=264 \rightarrow 5280+264=5544$

It immediately becomes clear that this method takes too long and can turn into quite a complicated mess, because you have to quickly combine non-trivial multiplication and addition. With this method you have to memorize the problem, then the outcome of the first multiplication, then do the second multiplication and remember all outcomes in order to add them together. We could think of every '$=$' sign as a mental step.

2) The second method comes from a branch called Vedic math. On paper this way of multiplying looks more complicated but with slight practice it becomes apparant that it's much quicker. It works like this:

• First multiply the two right most digist with eachother, yielding:

$3*8=24$

• We carry the $2$ and the $4$ is the last digit of our final answer. We then do a cross multiplication where we multiply the right digit of the second factor by the left digit of the first factor and vice versa, and add the outcomes together (not forgetting the carried $2$), yielding:

$8*6+8*3+2=48+24+2=74$

• From this we see that our second to last digit is also $4$, and we will again carry the $7$. For the last step we multiple the left digit of the first factor with the left digit of the right factor (not forgetting our carried $7$). This gives:

$6*8+7=48+7=55$

• We now know that our first two digits are $5$ and $5$, yielding our total answer of $5544$

Again, on paper this looks like many more steps than the brute force method but its much easier to keep track of the things you have to remember for the final answer.

3) An optional third method could be similar to the above method but instead of doing the cross multiplication as a second step, we do it as our first step and then do the original first step. I would imagine opinions on whether this is really quicker are divided but it helps with limiting the amount of calculations to remember.

4) Finally, the standard method could be to round up one of the factors to the nearest ten multiple and then subtracting whatever excess you added. In this case that would yield:

$63*90=5670 \rightarrow 5670-2*63=5544$

This method becomes much more complicated however when we try to compute something like $44*86$ because rounding to the nearest ten leaves us with much more excess.

Maybe the answer to my relatively broad question is simply: "fast multiplication comes from experience". However, I'm very curious to hear any other methods that are out there. Apologies for the long post but I hope I clarified my thought process enough.