# What is the fastest way to multiply two digit numbers?

I been playing different math games on my Android lately (for example: Math Cruncher).

I've noticed that I'm unable to quickly (under 7-8 seconds) multiply two digit numbers (i.e $18 * 17$). So my question is, what is the fastest way to multiply two-digit numbers?

• It helps to know perfect squares (so here you have $17^2+17$) and also use rounding tricks $17(20-2)=340-34$. – ShawnD Mar 5 '12 at 14:00
• When I do multiply it out, I find it faster to start with the most significant, here 10*10=100, 10*8=80 makes 180, 10*7=70 makes 250, 7*8=56 makes 306. It seems easier to keep track of where you are and (in other settings) can stop when you have enough figures. – Ross Millikan Mar 5 '12 at 15:13
• Some of the stage calculators were reputed to have memorized the table that far. Surely there is nothing faster. – Ross Millikan Mar 5 '12 at 15:21
• Algorithmically, the fastest way to multiply two large numbers involves a discrete fourier transform... but I doubt that's what you were looking for :) (also it's overkill for only two digits) – BlueRaja - Danny Pflughoeft Mar 5 '12 at 18:41
• ^ Schonhage-Strassen runs in ${\cal O}(n \log n \log \log n).$ If $n = 7,$ that's about $10$ operations. Not sure about the constant coefficients and low order terms though. /sarcasm :) – user2468 Mar 12 '12 at 21:20

The way I learned it:

$$\begin{array}{rrl} & 18 & \\ \times & 17 & \\ \hline & 156 & (10\times 10 + 8\times 7) \\ + & 70 & (10\times 7) \\ + & 80 & (10\times 8) \\ \hline = & 306 &\\ \hline \end{array}$$

In general:

$$\begin{array}{rrl} & {\rm AB} & \\ \times & {\rm CD} & \\ \hline & {\rm ACBD} & \text{(pad 0 if neccesary)} \\ + & {\rm BC}0 & \\ + & {\rm AD}0 & \\ \hline \end{array}$$

This is just dividing it into polynomial: $(10+7)\times (10+8)$. And it allows me to visualize easier than the traditional method.

The absolute fastest way to multiply two-digit numbers is to already know the result, by having learned the two-digit multiplication table just like you learned the one-digit multiplication table. Of course it means a lot of work up front.

My method for multiplying some two digit numbers is not going to work well in all cases, and has the draw back that you must memorize all the squares. But, I present it just as another trick you can learn. I +1'd FiniteA's answer as that seems like a nice way that works well in just about any case, and takes very little memorization or skill, i.e., it boils it down the simplest parts.

I would like to point out that my method is probably faster when the two numbers are relatively close together and when their difference is an even integer.

Using the fact that $(x - a)(x + a) = x^2 - a^2$, if you memorize perfect squares, you can do some multiplications pretty quickly.

Example 1: $24 \times 26 = (25 - 1)(25 + 1) = 25^2 - 1^2 = 625 - 1 = 624$

Example 2: $24 \times 27 = 24 + 24 \times 26 = 24 + 624 = 648$

Example 3: $24 \times 32 = 28^2 - 4^2 = 784 - 16 = 768$

1) Practice.

2) Practice some more.

3) ????

4) Profit!

Here are the things to practice:

(1) To compute $x^2$: use the identity $x^2=(x+a)(x-a)+a^2$, with $a$ chosen to make $x+a$ as round as possible. This is especially fast for numbers near to $50$ or $500$ or $5000$ and so on. I can do squares of numbers near $500$ in about 2 seconds this way. Example: $46^2=50*42+4^2$, further simplified if one simply remembers $50^2=2500$ and therefore that $50*42=2500-400$.

(2) In general, round up or down to the nearest multiple of $10$ and then correct: for example, compute $93*42=100*42-7*42$. This is especially useful if you don't need the exact value---then you get a good approximation very quickly. The one choice you have to make here is which of the two numbers to round, and you should do this to maximize the resulting simplification. In the example I chose, it's better to round $93$ up to $100$, since multiplying by $100$ is slightly easier than multiplying by $40$.

(3) As you begin doing mental arithmetic with larger numbers, you will realize that the primary obstacle is not speed but space: you will run into the problem that you cannot reliably store more than a few digits in your head at a time. To overcome this, you will need a mnemonic. One relatively painless way is described in Art Benjamin's book "Secrets of Mental Math: The Mathemagician's Guide to Lightning Calculation and Amazing Math Tricks": turn numbers into phrases, poems, stories or songs!

• For the particular example you mention, I would do it as 20*17-2*17. – Stephen Mar 5 '12 at 13:33

The mental calculators' trick is to start from the left :
$$\begin{array}{r} 17\\ \times 18 \end{array}$$
$$\begin{array}{c l} \text{'cross' computation} &\text{ partial result}\\ 1\cdot 1=1 & 1\\ 1\cdot 8+7\cdot 1=15 & 25\\ 7\cdot 8=56 &306 \end{array}$$

Another example $$\begin{array}{r} 237\\ \times 543 \end{array}$$
$$\begin{array}{c l} \text{'cross' computation} &\text{ partial result}\\ 2\cdot 5=10 & 10\\ 2\cdot 4+3\cdot 5=23 & 123\\ 2\cdot 3+3\cdot 4 +7\cdot5=53 &1283\\ 3\cdot 3+7\cdot 4=37 & 12867\\ 7\cdot 3=21 & 128691\\ \end{array}$$

The same trick (working from left to right) is used for other operations. Neat and fast!

There is some very nice videos on this subject on you tube. Here is one of them:

Fast Math Tricks - How to multiply 2 digit numbers up to 100 - the fast way!

I break one of the number down into 10's, so this would break down into 3 simple multiplications.

E.g.

\begin{align} &18 \times 17 = 18 \times 10 + 18 \times 7\\ &18 \times 7 = 7 \times 10 + 7 \times 8. \end{align}

On this logic $$18 \times 17 = 18 \times 10 + 7 \times 10 + 8 \times 7.$$

Takes a few attempts to get the pattern in your head but works for me!

You guys are making this way too hard. Just learn one times two digit inside and out, then do this. 69 * 89. Go 6*89 = 534. Add this to the first two digits of 9*89 = 801. Do it: 534 + 80 = 614. Glue the 1 on to the end. We get 6141. This has to be the best way since I'm a low-life chemist but can do them in seconds (no pencil or pen). Join the club now, my last offer.

If we want to calculate $a\times b$ and $a$ and $b$ are close to a round number $N$, then we can write:

$$a\times b = (N+a')\times(N + b') = N^2 + N(a'+b') + a'\times b' = N\times(a +b')+a'\times b'$$

In this case, taking $N = 20$ yields:

$$18\times 17 = 20\times(18-3) + 6 = 306$$

More examples:

$$984\times 993 = 1000\times(984-7) + 16\times 7 = 977112$$

$$\begin{split}9876\times 9913 &= 10,000\times(9876-87)+ 124\times 87 \\&=97,890,000+100\times(124 - 13) - 24\times 13\\ &=97,901,100 - 20\times(24-7)+4\times 7\\ &=97,901,100 - 340 +28\\ & = 97,900,788 \end{split}$$

## protected by Zev ChonolesSep 7 '16 at 21:19

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