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Speed Arithmetic – Multiply by Eleven
Introduction · Development · Rule for multiplying by eleven (harder version) · Rule for multiplying by eleven (easy version) · Carrying a one · Exercizes · Notes · Recommended Books
Here we’re going to show a wonderful method of doing arithmetic quickly – one that uses algebra to simplify multiplication, so that in fact the multiplication is not much more than addition. Worked out in the 1940’s by a Russian math teacher named Trachtenberg while imprisoned in the Nazi concentration camps, the method has been taught successfully to math-challenged children for more than 50 years now. We will take the basic idea, algebraic assists of basic mathematics, and show how to apply it in a variety of cases. The reader 0 will eventually see that this is no trick at all, rather a straightforward application of algebra. The method is very effective for any person who finds math difficult; the only requirement is (as usual) lots of practice.
One (1) is a special number in multiplication. It is called the identity element because any number multiplied by 1 is itself:
1 · 1 = 1
2 · 1 = 1 · 2 = 2
3 · 1 = 1 · 3 = 3
…
437265984102 · 1 = 1 · 437265984102 = 437265984102
…
This goes on forever.
Ten (10) is another special number in multiplication because we are used to using base ten arithmetic 1 , also known as ordinary arithmetic or just arithmetic. The effect of multiplying by ten is to move all the digits one place to the left, or to add a zero on the right:
1 · 10 = 10
325 · 10 = 3250
9873645 · 10 = 98736450
This will work with any natural number 1, 2, … and will also work with decimals as well (although we’ll leave that subject until a later time). Since
11 = 10 + 1
We can multiply by ten plus one instead of multiplying by eleven. In other words, the above two simple facts about multiplication can be combined to create a rule for multiplying by eleven. We’ll first state our rule "the hard way", then find a shortcut that does the same thing and use the shortcut instead.
Here is the first, "harder" version of the rule:
Rule for multiplying by eleven (harder version):
Multiply the number by one, multiply the number by ten, add the results.
Example
Multiply 234 by 11.
Solution
Multiply the number by one:
Multiply the number by ten:
Add the results:
Example
Multiply 4231 by 11.
Solution
Multiply the number by one:
Multiply the number by ten:
Add the results:
The method is easy, although at three steps, a bit labor-intensive. Let’s take a closer look at exactly what is happening here. Suppose we take a very simple example, like multiplying 43 by 11:
Multiply the number by one:
Multiply the number by ten:
Add the results:
How did we get the digits for the answer? In order from the least significant digit (the one on the far right) to the most significant digit (the one on the far left), we did this:
Notice that the zero in the last calculation for the "4" digit was implicit, meaning that there was just a blank space in front of the 43, so we actually read the 43 as 043. Putting a zero in front of a number like this does not change the number. In fact, we can put as many zeros in front of a number as we like without changing the number.
If we write the above three computations for the digits next to each other (note – this is not multiplying them – we are writing the numbers beside each other just like we do when we write a number with several digits), we get:
Take a moment to match each addition in parentheses to the additions above to assure yourself that we did the same thing. What we have just done is given each addition a place value. Now, if we ignore the zeros, we get:
So here’s what we did: first, we just brought down the 3 (least significant digit). Then, we added 3 to the next digit in order beside it, 4. Last, we just brought down the 4 (most significant digit). This might be clearer if we do this step-by-step. Let’s write the multiplication problem (not yet solved) like this:
We left a blank space in front of the 43, which we now know contains an "invisible" zero. Now, do the first step: bring down the 3 (the least significant digit of the multiplicand, or number being multiplied)
Next, add 3 to the next digit, 4, and write the result, 7, next to the 3 we already wrote down:
Last, bring down the 4 next to the 7:
What if we have more digits? In this case, the middle step (adding digits of the multiplicand that are next to each other) gets done several times. Let’s try multiplying 52314 by 11:
First step: bring down the number on the far right:
Next, add 1 to 4 (add the neighbor, 1 + 4 = 5) and write down the sum:
Do the same with the next two neighbors (3 + 1 = 4):
Do the same with the next two neighbors (2 + 3 = 5):
Do the same with the next two neighbors (5 + 2 = 7):
Since the 5 has no neighbor on the left, we can just bring it down:
We are done. We have found that 52314 · 11 = 575454.
We are now ready to state the rule for multiplying by eleven "the easy way":
Rule for multiplying by eleven (easy version):
Example
Multiply 712536 by 11.
Solution
Set up the multiplication as we have done above:
Bring down the digit on the far right:
Add the neighbors 3 and 6 (3 + 6 = 9):
Add the neighbors 5 and 3 (5 + 3 = 8):
Add the neighbors 2 and 5 (2 + 5 = 7):
Add the neighbors 1 and 2 (1 + 2 = 3):
Add the neighbors 7 and 1 (7 + 1 = 8):
We have now run out of neighbors, so we just bring down the digit (7) at the far left of the multipicand:
We are now done. We found that 712536 · 11 = 7837896.
It might seem that this is a very long process just to multiply two numbers, but in fact, when written down on paper, only one group of numbers is needed – the end result looks just like the last step above. The digits in the product (written below the underline) are written down one at a time, starting with the right side, until the multiplication is done.
So, the actual thing the reader writes down looks like it was done in "one" step.
Example
Multiply 54090632 by 11 in "one" step.
Solution
This is how the calculation looks when you write it on paper. We did each step in turn as above. Try it yourself and see that you get the same product.
Up to now, the neighbors we have added always totaled nine or less, so we didn’t have to carry anything. One of the nice things about this way of multiplying numbers is that you don’t have to carry anything bigger than 1. Let’s look at an example where you have to do a carry.
Example
Multiply 67 by 11.
Solution
Set up the multiplication as before:
The first step is exactly the same as before, write own the digit on the far right:
Now things are a little different, since 6 and 7 add up to a number larger than 9 (6 + 7 = 13). Write down the digit at the far right of the sum (the 3 in this case) and carry the 1:
Note where the carried 1 is put – on the "high" side of the next digit. This is deliberate.
The last step is also a bit different because of the carried 1. Instead of just writing down the 6, we have to add the 1 to it (1 + 6 = 7):
As before, we’ll really do this all in one step on paper. Let’s take a look at another example step-by-step first. Notice carefully where we put the carried 1’s.
Example
Multiply 8697 by 11
Solution
First, set up the multiplication:
Write down the digit on the far right:
Add the neighbors 9 and 7 (9 + 7 = 16), write down the right-hand digit of the sum (6) and carry the 1 (again notice where we put the carried 1 – it is now between the digits it will be added to):
Add the neighbors 6 and 9 and the carried 1 (6 + 9 + 1 = 16). Since this is also a two digit number, again write down the right-hand digit (6) and carry the 1 again:
Add the neighbors 8 and 6 and the carried 1 (8 + 6 + 1 = 15). Since this is a two digit number, write down the right-hand digit (5) and carry the 1 again:
The last digit on the left (8) needs to have the carried 1 added to it (1 + 8 = 9), then we write down the sum:
So 8697 · 11 = 95667. As you can see, doing the carry does not make the multiplication much harder. We can now try a much longer example, this time done in a single written step.
Example
Multiply 4986710279587 by 11 in "one" step.
Solution
Like all mathematics, doing this method with speed and accuracy requires practice. Once the problems here are done, additional problems can be invented and checked with a calculator (warning: most hand calculators are limited to about 9 to 12 digits of display at a time – our method works for any number of digits). We gave answers for all problems as a check – getting the answer right is as important as the method. After completing the exercizes below, the reader is encouraged to just write down about fifty more numbers, up to about eight digits, and work the method on each one. Check with a calculator. After about ten problems are worked, the reader should find that the problems become easier, so that the next ten take much less time. Practice enough so that you can multiply a ten-digit number by eleven in ten seconds or less – you’ll be surprised at how easy it becomes.
This method is not compatible with the standard, table-based method taught in U.S. schools and needs to be completed in order to be worth the effort. In light of the time required to write these articles, we provide a link to Amazon books for the original 1960 book that describes the method (see below).
Note that in computer science, base two (2) or binary, base eight (8) or octal and base sixteen (16) or hexadecimal arithmetics are commonly used – more on this in another article. We will always use base ten arithmetic unless we explicitly state otherwise.
The original, 1960 book that first brought Dr. Trachtenburg's work to the world's attention. Well written, brief, with plenty of examples.
An inexpensive Dover book that covers the same kind of topics. I haven't read this one (yet), but I have many other Dover titles, and am always happy with them.
(c) Phil Otken 1996-2004. All Rights Reserved except where explicitly noted.