# 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.

• Yes, all of those. Squaring a number ending in 5 can be useful e.g $35^2$ can be done as 3*4 = 12, put 00 on the end = 1200 then add 25 = 1225. Combine with difference of 2 squares so 33*37 =$35^2$ - $2^2$ = 1221. – Paul Nov 12 '18 at 13:01

Maybe the answer to my relatively broad question is simply: "fast multiplication comes from experience".

I really think it is.

All your methods work, and from experience, you'll know when to use ease of these methods.

On the other hand, here is a method you didn't use, and works pretty well in this case: [prime] factors decomposition.

You have to multiply 63 by 88.

$$63 \times 88 = 63 \times 11 \times 8 = 63 \times 11 \times 2^3$$.

$$63 \times 11 = 693$$

Therefore

$$63 \times 88 = 693 \times 2\times 2 \times 2 = 1386 \times 2 \times 2 = 2772 \times 2 = 5544$$

I favor this method anytime I have to multiply by powers of 2 (2,4,8,16,...), because "$$\times 2$$" is an operation your brain is used to do from really young age. Doing it several times can sometime be quicker that doing a one shot "$$\times 16$$" or "$$\times 32$$".

Moreover, here, when you're done with the powers of 2, you're left with 11, and "$$\times 11$$" happen to also be a really easy operation on 2 digits numbers.

• The method you mentioned is indeed new for me, so thank you for bringing it up. I wonder if it is actually worth studying multiple methods to the point where you can use them effictively, rather than mastering a single method though. However, I've been playing with the Vedic math method for quite some time now and feel that I am starting to hit a barrier in terms of speed. Eventually you will be able to execute everything at maximum speed and the only thing slowing you down is the amount of steps you must take. I guess it could also vary per person. – Charlie Shuffler Nov 12 '18 at 13:28
• I also think it vary form one person to another. And I'm really not into fast mental multiplication, so I'm not going to teach you what is good for you. I just wanted to share in case you find it usefull. – F.Carette Nov 12 '18 at 13:34

For me, it depends on why I am doing the problem and the accuracy I want. If I want an exact answer I work from left to right, doing the most significant multiply first. I will alter your example to $$63*87$$ so we can tell all the digits apart. I would first do $$6*8=48$$, then $$3*8=24$$ and add that on one digit right to get $$480+24=504$$, then $$6*7=42$$ and add that in $$504+42=546$$ and finally $$3*7=21$$ and add that in one to the right getting $$5460+21=5481$$. If lower accuracy is acceptable, you can stop part way through (more important if there are more digits-I can keep track of $$4 \times 3$$ this way when in practice).

For approximations I will often round and correct, so I would do $$63*88\approx 60*(1+0.05)*90*(1-0.02)\approx 5400*(1+0.03)\approx 5550$$

Knowing more facts helps. If you know that $$3*37=111$$ you can multiply $$63*37=21*111=2331$$. If you know $$7*11*13=1001$$ that helps a lot when it comes up. It is said that some of the stage calculators just know the multiplication table up to $$99*99$$

• +1 for $7 \times 11 \times 13$. – Ethan Bolker Nov 12 '18 at 21:58
• Interesting methods, but are they really feasible to perform if someone would ask you ''what's $ab*cd$", and you would have to compute it without them repeating the answer? Ofcourse when writing down a problem on paper it becomes much easier to see patters or simplifications on the problem but this becomes much more difficult when making notes isn't allowed. Anyway, thank you for your methods! – Charlie Shuffler Nov 13 '18 at 6:39
• Absolutely this is feasible if you practice. I would think of the factor $1+0.05$ as $+5\%$. As I said, I have done more digits than $2 \times 2$. – Ross Millikan Nov 13 '18 at 15:24

I am hoping that an example will be enough to show the general pattern. With practice, this is very doable.

If your memory is good, you can multiply a two-digit number times a two-digit number in your head. for example, to do

$$\begin{array}{r} 56 \\ \times 78 \\ \hline \end{array}$$

$$6 \times 8$$ is $$48$$. Place the $$8$$ and remember the $$4$$.

$$\begin{array}{r} \color{red} 4 \phantom{0} \\ 56 \\ \times 78 \\ \hline 8 \end{array}$$

$$5\times 8 + 7\times 6 + 4 = 40 + 42 + 4 = 86$$ Place the $$6$$ and remember the $$8$$.

$$\begin{array}{r} \color{red}{84} \phantom{0} \\ 56 \\ \times 78 \\ \hline 68 \end{array}$$

Finally, $$5 \times 7 + 8 = 35 + 8 = 43$$. Place the $$43$$.

$$\begin{array}{r} \color{red}{84} \phantom{0} \\ 56 \\ \times 78 \\ \hline 4368 \end{array}$$

If two digits in a row or column are the same, you can take a shortcut. For example, to multiply $$63 \times 88$$.

$$\begin{array}{r} 63 \\ \times 88 \\ \hline \end{array}$$

$$3 \times 8 = 24$$. Place the $$4$$ and remember the $$2$$.

$$\begin{array}{r} \color{red}2 \phantom 0 \\ 63 \\ \times 88 \\ \hline 4 \end{array}$$

$$(6 + 3) \times 8 + \color{red} 2 = 74$$. Place the $$4$$ and remember the $$7$$.

$$\begin{array}{r} \color{red}{72} \phantom 0 \\ 63 \\ \times 88 \\ \hline 44 \end{array}$$

$$6 \times 8 + \color{red} 7 = 55$$. Place the $$55$$.

$$\begin{array}{r} \color{red}{72} \phantom 0 \\ 63 \\ \times 88 \\ \hline 5544 \end{array}$$

• Yeah, this is exactly the second method I described. Perhaps I was too vague with describing it though! My problem with this method is that, when you are not allowed to write things down on paper, you have to remember the digits you already placed while doing new calculations. Often times I wil have forgotten my last placed digit when I arrive at the final multiplication, unless I repeat my first step really quickly halfway through. This is exactly where I make ''useless'' mental steps. This problem most likely comes from too little experience on my part though. – Charlie Shuffler Nov 13 '18 at 6:41

To answer this question you must make a very large number of assumptions, few of which are testable. For instance, it is conceivable that mnemonists can remember the entire multiplication table. Alternatively, often the first step is somehow classifying the problem (even times even, or sum of digits dividable by 3, etc.) and it is extremely difficult to know how many steps are being used. Moreover, some problems are "trivial" ($$20 \times 20$$ or $$10 \times x$$ for any $$x$$) and you won't perform them using one of the proposed tricky techniques. As such, will you quantify the "expected" number of operations?