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This is the story of the late Jakow Trachtenberg, and how to scam people using your mathematical tricks.

Updated on March 17, 2011

Jakow Trachtenberg

As an adult with kids in school, I found it terribly frustrating that school mathematics seemed to indoctrinate the kids into one way of thinking. In any problem solving scenario, there are multiple methods and several approaches. Mathematics should be a creative exploration, not to follow a rigid procedural dogma. Yet when trying to teach my kids, I repeatedly met with a barrier, "But the teacher says we HAVE to do it this way." and "If I don't show all the steps then I won't get any marks."

In this article, I hope to demonstrate that there are multiple ways of solving even simple arithmetic problems. Some methods work better than others in certain situations, and for certain people. No one way is best. Yet we are usually forced to follow a procedure rather than explore a landscape.

Herein is a story of the late Jakow Trachtenberg. He was a Russian who developed methods of speed arithmetic entirely without the aid of pencil and paper while in a concentration camp. It's amazing what the mind is capable of when put to the test.

He later perfected his methods and taught children who considered themselves useless at maths. These children were soon able to demonstrate amazing feats. For example, how would you mulitply 2392347834598 by 11? Can you do this in one hit, just writing down the answer, and then be able to check the answer?

This is possible. Consider the product 11 x 11. For small numbers – you can say, “one plus one is two, put it in the middle of the first number, giving 121.” There is a slight modification when using numbers who's digits add to more than ten, like for 84 x 11. In that case, you say, “8 + 4 is 12, answer is 8 twelvety 4”.... but of course, except for people like J.R.R. Tolkin, there are no numbers in the form of twelvety. So you put ten of that twelvety onto the 8, giving 9, leaving 2 for the middle digit, and the answer is 924.

You can check the answer too, and should do so for every calculation because it is so easy to do. Try this... to check 84 x 11, you can add 8+4 and do it like this 8+1=9 – forget the nine – we “cast out the nine”, and what is left is 3. Then for 11, one plus one is 2. So the new product, after casting out the nines, is 3 x 2 = 6. Six is our checksum, and the product of 84 and 11 must also contain digits which make a checksum of 6. Lets test it... Our answer was 924. Now, 9 can be cast out immediately. Just mentally cross it off. 2+4 = 6... and that is our checksum. So our answer is probably correct. I have to say “probably” because there are many other numbers that also have 6 as a checksum, but not too many are close to the correct answer, so the likelihood is that our sum is correct.

Here is another easy example...

is 9823499 * 7276342 = 23882773?

Before proceeding please get familiar with casting out nines.

Cast out all obvious nines, and test

((8+2+3+4)) + (( 4 )) = ((3+7+3))

((8)) + ((4)) = ((4)) , answer is NO, the sum is proven wrong.

This time we KNOW for certain that the answer is wrong because the checksums are not equivalent. This is because there is only one correct answer, and for that, the checksums must be correct, so it follows that an incorrect checksum proves that the answer is wrong.

What are all those brackets for?

In mathematics, we can make up notation to cut down on the amount of words that we need to use. Once the notation is well understood, it makes it easier to find patterns and relationships. It also removes ambiguity- that's is a problem in general language where it's not clear how to interpret a single statement. In mathematics, we need to be very precise. I just invented the (()) notation to indicate that we are working out a checksum. You see, the checksum for 436 is 4+3+6 which equals 12 and the checksum for 12 is 1+2 = 3. I wanted to be able to write it like this: 436 = 12 = 3, but of course, that is ambiguous because we know that 436 does not equal 12, and we know that 12 does not equal 3. However, the checksum of 436 = 12. So I say, ((436))=((12)).

In maths, you can make your own rules. But they have to be unambiguous and consistent and give the right result.

A chain of reasoning

Now we can easily write down a chain of reasoning for casting out nines without making mathematicians cringe about ambiguity. Here is an example.

Find (( 998349879873463487230345873498 ))

= (( 834878734634872334587348 )) ignore all zeros and nines

= (( 8348787346348334587348 )) strike out all 7s paired with 2s

= (( 8348787346348334587348 )) there are a few 8s and 3s replace 3 by 21

= (( 8214878721462148212145872148 )) now there are more pairs that add to 9

= (( 2446424524 )) we got rid of 8 paired with 1 and 7

paired with 3

= (( 24464224 )) and now 5 paired with 4 goes

= (( 231316312231 )) 4 becomes 31

= (( 6361633231 )) 33 becomes 6, and 21 becomes 3

= (( 1231 )) remove 6 paired with 3

= (( 7 )) this is the check sum.

What about the first big multiplication by 11 ?

2392347834598 x 11 ?

Well, when we do this the traditional way, there is an obvious pattern.

 X	   11
   23923478345980 +

Notice how we are mostly operating on two digits next to each other? The rightmost 8 gets added to a zero (one that we can imagine is there), then the 9 gets added to the 8, and the 5 gets added to the 9, and so on. Sure, there is a bit of carrying to do, but the task is focused on simple little steps and there seems to be a little trick to do when you run out of digits – in the same was that there is the trick when you start by imagining a zero to the right of the rightmost 8. Once you know the key, then it's not significantly more difficult to multiply large numbers compared to small ones.

This example gives a little clue about Trachtenberg's methods. He found ways to do arithmetic that are simple compared to the traditional methods. His methods make it possible to add a thousand numbers and more without counting higher than 11. There are no multiplication tables, and no division. He theorised that children have difficulties with arithmetic because of the clumsy methods that are traditionally taught, and illustrated this by setting out to teach those who struggled at school. He found that these children's self esteem improved and they made progress in all areas of study.

The Trachtenberg system chases away the fear and sometimes disgust for mathematics that many children learn.

There are some definitions to learn in his system. He describes numbers in new terms so that it is easy to perform the calculations. Here are the definitions. They are not hard, and they are reasonably named.


d = digit currently worked on.

ds = digit sum.

n = RIGHT neighbour.

o = odd.

e = even.

pp = pair product.

t = tens digit.

u = unit digit.


I sometimes use a star * to mean “times” because it does not get confused with X for the letter X as often used in algebra.

Even-numbers can be divided by two without a remainder. That should not be a surprise. But in this system, odd numbers, when divided by two also have no remainder. That might be a surprise unless you are aware of what is called “integer division”. Working only with integers is common for computer programmers. An integer includes all the counting numbers, { 1,2,3... } , zero, and all the negative counting numbers, { -1 , -2 , -3 ... }. There are no fractions, no decimal points and so on. Therefore, an integer division of 9 by 2 is the same as 8 by two because you lose the “1” that is left over.

So half of 1 is 0 for this integer arithmetic.

We are making progress already !

To multiply by 11, we use definition “n”

Here is the rule:

To multiply by 11, add the neighbour...

Let's try it.

23 * 11

Imagine 0 is 3's right neighbour.

3+0 = 3

write “3”


write “53”

Is that it? But don't we need the “2” in there as well. How ? Well, there is also a zero in front of “23” - can you “see” it?

023 * 11

(it's the same thing... but we need it for our system.)


write “253”

Check the answer by casting out nines...

23 * 11 = 253

((5)) * ((2)) = ((1))

((10)) = ((1))

((1)) = ((1))


We can also see how the trick of adding the digits, and inserting into the middle works.

Let's try one where the digits are bigger, and therefore add to more than nine.

78 * 11

think of 0780




Reading upwards,

the answer is 7,tentyfour,8

But tenty is silly, so we say


Check it...







OOPS! What is wrong? You may have already seen the error. But lets check our working.

Ah, yes, there is the problem... 7+8 is 15, not 14. So we are one digit out, and the answer

should be 7,tentyfive8 which is 858. Do we need to go through all the steps in the checksum again?

No. because our correction was by +1, and our checksum was one too small on the right hand side.

“eleven” is a pretty nice number. It's only one more than ten, and multiplying by 10 is easy, so it's not surprising that mulitiplying by 11 is easy. But in mathematics, things that work for one number will work for other numbers. The catch is that the particular numbers involved work easiest for certain methods.

A diversion

Let's take a diversion from Trachtenberg for a while, and look at multiplying two-digit numbers that end in 5.

05 x 05 = 25

15 x 15 = 225

25 x 25 = 625

35 x 35 = 1225

45 x 45 = 2025

It's not a coincidence that all these answers end in “25”, all squared numbers that end in 5 end in 25, and we can prove it.

You might also notice that the other digits are related to the digit in front of the five.

0 x (0+1) = 0

1 x (1+1) = 2

2 x (2+1) = 6

3 x (3+1) = 12

4 x (4+1) = 20

and so on. So to square a number ending in 5, for example 75 x 75

(think: 7+1 = 8 and 7 x 8 = 56 ... the answer is 5625)

Why does this work?

Let's see if it works for any number ending in 5. We don't specify that number, so let's call him Fred.

Fred is in the tens column isn't he? So he is ten times his digit-value. We say 10Fredsworth.

And the number that we are considering is 5 more than that... so another way to say the

whole group of numbers that end in 5 is:

10Fredsworth + 5

What would 10Fredsworth bananas look like?

If Fred was the number 4, then 10Fredsworth bananas is 40

and 10Fredsworth + 5 is 45

We could do this for any value of Fred. Our notation is watertight.

And since it's watertight, we could just as well used Fishe'sbum. But it's such an ugly thought.

So what is 10Fredsworth of bananas TIMES 10Fredsworth of bananas ?

That would be like looking at 10Fredsworth rows of bananas, each row containing 10Fredsworth bananas. This could be a good feast for a monkey if Fred was a decent sized number.

What about ( 10Fredsworth + 5 bananas ) rows of ( 10Fredsworth + 5 bananas ) ?

(noting that this is how we could say 45 squared, if Fred had the value of 4)

How do we multiply that?

I'm getting tired of typing 10Fredsworth. It's too long and I hope you don't mind if I just use 10F instead. Is that okay? Is it too big a leap to extend this to 10x? Can I say “10x means exactly the same as 10Fredsworth?)

Where were we? Ah yes, finding (10x+5) rows of (10x+5)

How do we do this? How can we write the problem down in the first place?

It doesn't really matter how we write it down, as long as we can understand the notation, and can teach others to understand it.

Let's not stray from worldwide convention, and write the problem like this:


In the same way that the close cuddle between Fred ( who is really 'x' now) and his 10 that represents his position in the ten's column, we don't bother writing down the multiplication sign. So 10x is a quick way to write 10 x x... and now you see why we do this. The x and the x get confused don't they? Which x is x, and which x is x? You don't know? I meant, which x is EX and which x is times? It's better just to drop the x and pretend it's there.

Ab means A TIMES b

15c means 15 TIMES c

Don't let this scare you. You are used to saying 15 dollars, not 15 times dollars. And you write it $15 not $x15, even though it means the same.


(10x+5)(10x+5) – how to do this?

Think of a field of bananas again.

If x was 2, then 10x is 20, and a field of 20 rows of 20 bananas is a good start.

But if we added 5 bananas on each row, and five more rows, then the total number of bananas is ... what? Let's look at this with a diagram...


Each dot is a banana in the original planting, while the + is for 5 more bananas per row, and five more rows.

The * happened because the + from the additional rows clashed with the + from the extras in the row. And also I wanted to point out that there are exactly twenty five of them. There is always twenty five of them, no matter what the original field was. If we add five rows and extend each row by five, then there is a little square of 25 bananas in every case. This is partly why any number ending in five when squared, ends in 25.

How many +s are there? You could count them, excuse me and I'll take a nap while you do it. Or we could say, 5 rows of however many bananas were originally in each row.... twice. (Once for the ones at the bottom, and once for the ones at the right hand of the field. In our particular case, that's twice of all of : 2 times 10, five times.

Finally, we can see another way to write


It's 10x time 10x ( all the dots )


5 * 5 ( ( all the stars )

plus twice 5 lots of 10 lots of x ( all the +s )

And we can write this much neater like this:

10x10x + 2(10x5) + 25

Mostly in the future, you will multiply the 2 and 10 and 5 and x to get 100x, and 10x10x as


that is:

100x2 + 100x + 25

So again, we see that little “25” all on it's own, to be added to some other number.

Let's look at the case of x=3, so that 35 times 35 is our question. The answer is 1225.

We see where the 25 came from as just stated, but what about the 12?

Well then 12 is really 1200 isn't it?

And if x is 3, then our equation solves to:

(100 times 3 times 3 )+ (100 times 3) + 25

Which is

(900) + (300) + 25

whoopee, 900 + 300 = 1200

and we see where the 12 comes from. And the two zeros are because of the multiplication by 100 in100x2 + 100x.

But we noticed also that to square a number ending in 5, we add 1 to the rest of the number then multiply it by that incremented number. How does this work?

It must be another way to solve

100x2 + 100x

or more specifically

x2 + x

because we don't really care about the easy multiplication by 100 – we can take it out for now and think-it back in later.

We are “postulating” ( that means sort of guessing ) that

x2 + x is the same as (x+1)(x + 0)

This (x+1)(x + 0) is almost the same form as

( 10Fredsworth + 5 bananas ) rows of ( 10Fredsworth + 5 bananas )

Except that it's 1 banana in the left, and 0 bananas in the right.

More bananas

Here are x bananas


(four of them)

Here are x rows of them:


and now we add one row, and none extra in each row:


See ?

Look how x is 4, and 4*4 is 16, (The number of dots)

and one extra row is 4 more (The number of +s)

so we can easily see that if 4=x,

x times (x+1) which is 4 rows of 5

is the same as

x2+x which is original 4 time 4..... and 4 more.

Mathematics is full of patterns like this. Many problems in the real world can be visualised with a field of bananas and ravenous monkeys, and even more can be visualised with geometric shapes.

Getting back to Trachtenberg

Getting back to Trachtenberg, let's first review his way of multiplying by 11 by practising on a long number.

This time, I will separate the digits with a space, so that we can use a little dot to represent when we need to consider a carry.

0 6 1 2 7 6 2 3 7 6 times 11
 6 -------- 6 + 0 = 6
.3 -------- 7 + 6 = 13
.1 -------- 3 + 7 + 1 = 10
 6 -------- 2 + 3 + 1 = 6
 8 -------- 6 + 2 = 8
.3 -------- 7 + 6 = 13
.0 -------- 2 + 7 + 1 = 10
 4 -------- 1 + 2 + 1 = 3
 7 -------- 6 + 1 = 7
 6 -------- 0 + 6 = 6

It's good to put that leading 0 in place until you are swift and competent with the method.

Let's cast out nines to check the answer.

(( 612762376 )) =

(( 616 )) =

(( 4 )) This is the checksum for the big digit.

The checksum for 11 is (( 2 ))

(( 2 ))x(( 4 )) = (( 8 )) This is the checksum for the product, let's compare it with the answer that we got.

(( 6740386136 )) =

(( 674 )) =

(( 71 )) = (( 8 ))

The answer checksum matches the product and the answer is probably correct.

With a little practice, multiplying any number by 11 becomes literally child's play. When you become good at working on a digit and its right neighbour in this way, then move on to the next method which works for multiplying by 6.

Multiply big numbers by 6.

You would think perhaps that multiplying by 6 is easier than 11 because it is smaller but 11 is just one more than 10, and multiplying by 10 and by 1 are easy. The 6 times table is more complicated.

But Trachtenberg found a way to multiply by six using only the skill of adding two single digits and also by dividing single digits by 2. Remember that we talked about integer arithmetic? Effectively throwing away the remainder? Well it's now that this comes in handy, and for good reason that will become apparent later.

When you can grasp

5 + 4 + 1 = 10


9 integer-divided by 2 is 4


3 is an odd number, while 8 is not... then you will be able to multiply any number by 6 without greater powers of calculation.

Let's do some practice first.

Think of 3, it's odd, start counting from 5
Think of 2, it's even, start counting from 0
Think of 3, it's odd, start counting from 5
Think of 5, it's odd, start counting from 5
Think of 7, it's odd, start counting from 5
Think of 9, it's odd, start counting from 5
Think of 4, it's even, start counting from 0
Think of 6, it's even, start counting from 0
Think of 8, it's even, start counting from 0
Think of 0, it's even, start counting from 0

Now we are ready for an example. Let's do 3214 times 6

The first time is hardest. You need to find a method of thinking about the digits and where to place them. I will describe the way that I settled on after a few practices. When you get the idea, you can invent your own version.

Write down the number.


Underneath, put a 5 wherever there is an odd number. Place a leading zero.

 5 5

On top, we do the integer division trick. But start with the 1 and above that halve the 4.

 5 5

Now above the 2, put integer-half of the 1 which is zero.

 5 5

Now above the 3, put a 1 because it's half of 2.

 5 5

Above the zero (you put it there as a reminder) put integer half of 3

 5 5

Now add the columns and you have the answer.

 5 5

Check by casting out 9s.

((19284)) = ((24))=((6))


((03214))((6)) = ((1))((6)) = ((6))

so we are correct. (most likely).

A longer example

See if you can find the error in this longer example.

 5 55 5 5 5 5 5 5 55 5 5  5

Here is the answer.


You may be puzzled about this integer-half thing. How is it that we can throw away the left overs and still get the right answer? What is this mysterious trick about adding 5 for odd digits? Actually, we are not really throwing away the odd left over 1 at all. It sort of gets back in there because of the “5” trick.

Think of it like this. If you take 10, and halve it, then you get 5. There are no great surprises there, I know, but if you integer-halve 9, you get 4. Double 5 and you get 10 again, but double 4 and you only get 8. Hold that thought, we need to demonstrate something else first.

When we multiplied by 11, we added the active digit to the one on the right. It was easy to see why we did this, because it's (10x) + x and to multiply by ten you shift the whole number to the left and replace the empty space by a zero.

1234 * 11 looks like this when you do it the conventional way:

 1234 +

You can see how we are really adding the active digit to the one on the right. So when we multiply by 12, we expect to double the the active digit and add to the one on the right. In fact, this is precisely what Trachtenberg does for multiplication by 12.

Conventional example:

1234 times 12
 2468 +

Trachtenberg example:

    |___________ To get this digit: (2*6)+5 plus a carry.

Now since 6 is half of twelve, we should see a closely related method as multiplying by twelve

In fact, to multiply by twelve, what you can do is first multiply by twelve, then divide the result by two. We can take any (even) number, and see if there is a digit-by digit way to divide it by two.

Try something simple first.


(Simple enough?)

divide the active digit by 2, we get 1.


same again – no complications. We get 11


Now it's harder. ½ of 3 is 1 ½

For the value 3, we turn it into 1, and then what happens with the ½ ? We need to think about what this ½ really means. We are in the 10s column, so a 1 in that column is worth 10 if there were a way to cram it into the column on the right, and therefore, ½ in this column is worth 5 if we cram it into the column on the right.

3  2
1½ 1 +

is the same as:

11 +
11 + 5 = 16 ( which is half of 32)

Try this out on a bigger number

1 6 5  5  3  3  5  2 7  4 6
½ 3 2½ 2½ 1½ 1½ 2½ 1 3½ 2 3

Which equals:

0  3  2  2  1  1  2  1  3  2  3
   5     5  5  5  5  5     5


0  8  2  7  6  6  7  6  3  7  3

It works, and we can skip the messy ½ business, and make a rule. To halve a big number,

write a 5 under every odd digit...

5 55555 5

Shift the 5s to the right.

 5 55555 5  

Integer-halve the digits on top...



5 55555 5

Add up.



5 55555 5



And that matches our answer before. At least some of the mystery has gone from the Trachtenberg system for multiplying by 6. We can foresee that to multiply by 6, the algorithm will be a short cut to the combination of the two algorithms for first multiplying by 12, then dividing by 2. His system combines the two steps into one.


The Jedi Mind-Trick

The Jedi Mind-Trick

It's time for another little diversion. Let's look at an easy way to subtract nine. Most people work out that you first subtract 10, then add one. But my brain sometimes gets a bit lost, forgetting to add the right way – not being sure of the answer, needing to re-work the logic. It's all a mind-trick. Here is the same information, but presented with a different slant.

To subtract nine, just read out the number, digit by digit, but when you get the the second-last one, take one away, and add it to the one on the right.

4322 – 9 =


8887236 – 9 =


For some reason, this is much easier, even though we are doing exactly the same thing as taking 10 away and adding 1 to compensate for taking 10 away.

There is a tiny complication when you get a 0 in an awkward spot:

376803 – 9 =


But you can read the original as “three seven six seven tenty three” and say

“three seven six seven ninety four” and get the answer right.

It's handy to be able to subtract 9. It's not not much harder to subtract 8, or 7 and it should work for adding 7,8 and 9 as well. Let's try a few samples.

636 + 9:

We say, “six”, think,”this is the second last digit, give one.” say, “four”, think,”Now it's pay-back time so take one”, say, “seven”.

That is a lot of explanation, but it is important to have this little narrative in your head while you create the answer. It's this chit-chat in the head that allows you to concentrate on the method, and remember things that need to be held for a short-time. You need to make use of what researchers call “short-term auditory memory”. Many people can do this quite well, but some are not so good at it, and there are alternatives. I'll talk about the different way's that the minds works as we develop these methods. Try this experiment:

Read this number, then turn away and recite it:


How far did you get? Most people find seven digits are comfortable, and beyond that is a challenge. Try this number:


That should be a lot easier. To find out what your threshold is for short-term auditory memory, try this test:

For each of these numbers – get someone to read them out and then you recite them backwards. When it's gets too hard then stop. That is your current number of items that you can reliably hold in short term memory. By the way, don't despair if you don't get far. It does not mean that you are dumb if you can't do well on this test because some extremely intelligent and productive people have poor short term memory. There are some tricks that you can use to improve. We will talk about this later.

With the first number: 776554432819, there are 12 items if you think an item is a single digit. However, by breaking these into groups, there are only fewer items:

77 65 54 43 1928

That might help you. But there are other tricks that you can use. Ironically, adding more things to remember can actually make it easier!

An old lady of 77 once told me that the best birthday she had was when she was 65. They could only fit 54 candles on the cake, she only blew out 43, and that was way back in 1928.

This works because it sets up links and relations from one item to the next. The story helps you to remember things about the next number in relation to the previous. I find it easy to picture a 77 year old woman talking about her past. It's just coincidence that the pairs of number that follow are getting smaller, but that is another observation that is weaved into the story and emphasised by the word “only”.

You can also make numbers more friendly by giving them personalities. There are lot's of ways to do this, and here is just one:

0	is an egg, an open mouth, a black hole or something that rolls.
1	is a snake, a very tall man, a piece of string or something 
        that can blast off.
2	is a swan, a little old lady, the captain's hook, or an item 
        that's broken.
3	is a caterpillar, a pregnant lady, a pair of handcuffs, 
        or something yummy.
4	is a horse, a man sitting down, a painter's easel, or 
        something slippery.
5	is a seahorse, a very fat kid, a coffee machine or happiness.
6	is parrot, a man in a wheelchair, a handbag or something unstable. 
7	is a camel, a school teacher, a walking cane or someone up-tight.
8	is an elephant, a big fat lady, a dog chew, or someone in a tiz.
9	is a lizard, an athlete, a leaf blower or something very fast.

776554431928 now takes on a new personality. There are several stories that we can make up about this number, and here is just one:

Once upon a time, (all good stories start like that)

there was a camel with a walking cane! He was owned by a man in a wheelchair who also kept a seahorse in a coffee machine. While sitting there, painting it, a caterpillar blasted off a piece of string very fast and broke everything. This put the man in a big tiz.

Yes. The story is very silly, but that is the very reason how it works as a memory aid. The silliness connects to our emotions, and the pictures that are created in out head make the numbers easy to remember because we have given the number personalities.

You might have noticed that each number was assigned something to do with an animal, some part of the body, an object and an emotion or an action. This makes it easy to create stories about people, animals, things and stuff that happens.

Let us examine the relationships between the numbers and the animals:

0 It looks like an egg.

1 A snake is the same shape.

2 Imagine the shape of a swan gliding along the water. It has the shape of a 2.

3 Turn your head to the right. This looks like a caterpillar with multiple legs.

4 Horses are angular creatures, so it the shape of a 4.

5 This is a stubby figure with a plump tummy – like a seahorse.

6 The shape of the 6 looks like a bird sitting on a branch. A parrot has more character than just thinking “bird”. Parrots can talk. That's useful in a story.

7 7 could be a giraffe but a camel has such a lot of character. They both have long necks.

8 8 is a big chunky number. An elephant is a big chunky animal.

9 This shape re-minds me of those lizards that run so fast that they can walk on water. I liked that.

Now it's your turn. You should make up your own relations between the digits, and the objects or people etc, and write them down somewhere. Alternatively, you may not like the relationships that I have chosen. In that case, you can make up your own. The useful part is to be able to assign definite personalities to each digit. You don't have to stick to one set of links either. When I was studying for MCP, the networking section demanded that we learn the rough contents of many of the standards that were written for computer communications. They all start with “802”, so that was no problem to remember. But there were sub-sections, like 802.11 and 802.3 and 802.5. These are for wireless networks, something called Ethernet, and something called token ring. You don't need to understand what these are, but it is useful for you to understand how I remembered the link between the numbers and the content of the sub-sections.

802.11. The 11 looked like two radio -towers. This is wireless communication.

802.3 The three looked like our caterpillar. I imagined little caterpillar walking down the wire – and this was good enough for me to make the link the the thing called Ethernet.

802.5 The 5 looks like a ring. Hence token RING.

Short-term memory is the ability to remember items in isolation, without links between them, for a short time – typically this is only a few seconds. Long term memory relies on putting these items into a story. This is called putting into context. It's the time-consuming, but usually fun part of learning, and its made more fun by employing silly memory tricks with parrots, rocket propelled caterpillars and painters in wheelchairs. The story that puts things in context can either be a made up one, like the one that we just did, or it can be the discovery of how information fits into the whole.

We need to get back to the adding and taking away of 7,8,9 and see how that fits into the diversion about short and long term memory.

We need short term memory to cope with special cases like a number that has a zero in an awkward spot – for example: 887508 where we need to push some new information into our memory, hold it for a short time, then use it later. In this case, its the conversion of the 0 into “tenty”, and the reduction of its digit to the left. The use of the silly word “tenty” aids our short term memory. All the while, we are accessing long term memory to get at the story about taking, and compensating a value from the last but one, and the last digit. You can imagine two lines of thought. These lines of thought are independent, and because different parts of your brain are dedicated to long and short term memory, it is possible to do these two tasks at the same time.

When we take or add 8, the juggling involves the number 2 because 2 is the difference between 8 and 10.

When we take or add 7, the juggling involves the number 3 because 3 is the difference between 8 and 10.

Luckily, perhaps by using our 10 fingers, most people can instantly see the difference between a single digit number and 10. Here they are listed:

9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7 8 9

I see these as two wedges like this:

If you do not instantly see the 3 as a mate for 7 , or a 4 as a mate for 6, then it's worth practicing so that you do get this ability. We can use it in many places during a mental calculation to aid the short term memory.

Try these – using the methods described:

534 – 8   MINUS! 8 is paired to 2, say 5,   think 3-1, say 2, think 4+2, say 6 526

976 – 8   MINUS! 8 is paired to 2, say 9,   think 7-1, say 6, think 6-2, say 8 968

456 + 9   PLUS!  9 is paired to 1, say 4,   think 5+1, say 6, think 6-1, say 7 465
456 – 9   MINUS! 9 is paired to 1, say 4,   think 5-1, 6+1, say 47 447

325 + 8   PLUS!  8 is paired to 2, say 3,   think 2+1, think 5-2, say 3 333

34523 – 7 MINUS! 7 is paired to 3. say 345, think 2-1, say 1, think 3+3 say 6 34516
99832 + 7

If you use this method on the last example, then it looks hard. Blindly following the method makes you want to take 3 from 2 in the last digit, which is negative, then you need to re-adjust the 3 that turned into a 4, back to 3 again. Although not wrong, it is an example where one particular method is not the best to use. So when you look at a problem, take stock of the situation, and decide which method is easy for you to apply. As you build up familiarity with more methods, it becomes easier to spot the best method to use at the time. In the last example above, it's easiest to just add 7 + 3 because the result does not affect the 10s column.

Getting good at adding and subtracting single digit numbers to and from double digit numbers is really handy for Trachtenberg's methods because most of the time you are doubling, halving, adding or subtracting active digits, and their neighbours. Knowing instantly the pairs of numbers compared to 10 is very useful.

We have practised multiplying by 11. We have practised multiplying by 6, and worked out how to multiply by 12. We have also noticed that dividing by two can be done sing the same sort of tricks, and that dividing by two in the same way give us reason to at least slightly understand that Trachtenberg's method of multiplying by 6 is similar to the method of multiplying by 12. We conclude the multiplying by 6 is the same as first by 12, then dividing by 2. Both these operations are easy, and Trachtenberg presents a method that automatically combines the two.

Here are Trachtenberg's definitions again.


d = digit currently worked on.

ds = digit sum.

n = RIGHT neighbour.

o = odd.

e = even.

pp = pair product.

t = tens digit.

u = unit digit.

Using this “notation” we can describe the methods that we have so far practised.

To multiply by 11:

d + n

and we could also reduce this to just


which means “digit sum”

To multiply by 12:

2d + n

To multiply by 6:

o: 5 + d + integer half n

e: d + integer half n

The colon after the o and the e is to indicate, “this is what you do” and you would read o: as, “For odd digits, this is what you do.” Also, e: is read, “For even digits, this is what you do.”

Let's try this last one again, this time using the notation as explanation.

024576 * 6

|||||+------ e:     d + integer half n = 6+0/2 = 6    (a)

||||+------- o: 5 + d + integer half n = 5+7+6/2=15   (b)

|||+-------- o: 5 + d + integer half n = 5+5+7/2+1=14 (c)

||+--------- e:     d + integer half n = 4+5/2+1= 7   (d)

|+---------- e:     d + integer half n = 2+4/2 = 4    (e)

+----------- e:     d + integer half n = 0+2/2 = 1    (f)

In (c) and (d) above, we had to remember to add the carry from (b) and (c). And in (f) we recall that the leading zero is required to make us finish the sum properly, and zero is an even number. The answer is 147456. Check this using checksums:

((24576)) * ((6)) =

((456)) * ((6)) =

((6))*((6)) = ((36)) = ((9)) = ((0))


((147456)) =

((1476)) =

((12276)) =

((126)) = ((9)) = ((0))

The answer is probably correct. Note that I keep saying “probably correct”. I explained why quite some time back, but you might wonder what are the chances of getting the checksum right for a wrong answer. A quick estimate to that question is “less than 1 in 10 times.” I can say this with confidence because there are 10 checksums, so on the face of it, if you guessed an answer at random, there is only a 10% chance that it is the right answer. But in practice, since you are using a method to get the answer where the method works mostly on pairs of digits, then the final answer is going to close even if it is wrong. Any single-digit error in the answer will certainly cause a wrong checksum, so we can say that casting out nines detect a single digit error to 100% accuracy. Of course we could make a multi-bit error, but it happens less often, and it certainly does not happen at random. Casting out nines as a checksum on you work is a better than 90% accurate way to detect errors.

I showed you how to group pairs of digits that add to nine, and to expand digits to make the pairs happen. There is another method that you could use to make calculating a checksum easier.

Take this number for instance.


9 and 0 go away

Pairs happen from outside to inside because they are in order. 8 pairs with 1, 7 with 3, 6 with 4, leaving 5.

What advantage does this give us? Let's try a large checksum problem and see.


Order them

((00 222 33333 44444 5 66 77 888 999))

We can forget the 9's and the 0s'

((222 33333 44444 5 66 77 888))

Now we have a choice. We can pair 8 and 2 to get ((1)) and we can pair things that add to ten like 7 with 3 and get ((1))

((111 11333 44444 5 66 ))

The 6s and two of the threes go out, and one of the 4s can be cast out with the 5

((111 113 4444 ))

4+4 is 8, so we can think of ((44)) as ((8)) and these are knocked out with two of the 1s

((111 3 ))

and this adds to ((6)).

A minus times a minus is a plus.

Multiplying is just repeated adding, so that 6+6+6+6+6 is the same as 5 x 6.

If you take a big number from a smaller one, then the answer is negative. 5 – 9 is -4.

We can put this concept into the real world by imagining that we have 5 dollars, but would like to buy something for 9 dollars, and therefore we need to borrow 4 dollars to get enough money. The borrow part is represented by the minus sign.

So far, this is good. But in school, we are (correctly) taught that a plus times a minus is a minus, and a plus times a plus is a plus, and a minus times a minus is a plus. How is this? What does this mean? Can we give an example with dollars?

Some examples just using numbers include:

+5 times +4 = +20 ----------(1)

-5 times +4 = -20 ----------(2)

-5 times -4 = +20 ----------(3)

In (1), if multiplying is repeated adding, then 5 times 4 is the same as (+4)+(+4)+(+4)+(4)+(+4)

or we can say 4 lots of 5 which is (+5)+(+5)+(+5)+(+5)

In (2), we can do the same thing for 4 lots of (-5) which is : (-5)+(-5)+(-5)+(-5) but how can we visualise (-5) lots of (+4) ?

(-5) lots of (+4) is like saying, “I've got five piles of money, each pile contains $4” I can spend one pile , then another, then another, then another and the final one. How much have I spent? I've spent 5*4 dollars. But the key word here is SPENT. I've spent this money, so I no longer have it. It's gone, negated, MINUSed from me. So MINUS 5 times $4 is MINUS $20.

Now we can see what MINUS $5 times MINUS $4 means. Now we have four I.O.U notes on the table from someone who has borrowed money from us. Each I.O.U note is for $4. If I SPEND an I.O.U note. Then I give it to the person who owes me money and I get money in return. This is why a minus times a minus is a plus. An I.O.U. Note is negative. It represents money that you don't actually have. SPENDing means giving away, and giving away an I.O.U. note to the person who owes you money results in a positive amount of money in it's place.

When you see this:

2(3+4) it means 2 lots of (3+4) which is 2 times 7 = 14

This could also work by multiplying the 3 and the 4 first, then adding the result:

2(3+4) = 2(3) + 2(4) = 6 + 8 = 14.

In general (that means for any set of numbers)

x(p+q) = x(p) + x(q)

So when you see something like this:


you can re-write it as

a(b) + a(-c)

but since a plus times a minus is a minus,

a(b-c) = ab-ac

We will use this result in the next section.

Multiply in your head.

Another trick for multiplying numbers is presented in a book by Bill Handley called “Speed Mathematics”. This particular trick is called the reference number method.

If both numbers are close to another number that is easy to multiply – like 100 for example, then it is trivial to do some impressive number-work in your head. You can multiply 104 * 106 for example just by looking at it. This is how to do it:

The reference number is 100

104 – 100 = 4

106 – 100 = 6

104+6 = 110

( by the way, 106+4 also = 110 – you can choose either since you end up with 110 in either case.)

110 * 100 = 11000

add 4*6 and you get 11024

Check:	((104))((106))=((5))((7))=((35))=((8))

Using some algebra, and especially the multiplying-out with negative numbers that we looked at in the previous section, we can prove why this works. Let's call the two numbers to be multiplied a and b. Remember that we can write this without the multiplication sign like this:


we pick a reference number and call it r.

Then the method is to take the number a and subtract the reference number. For this we get a-r.

We do the same for b to get b-r

Then we took a and added it to b-r ( or took b and added it to a-r )

That number is multiplied by r and added to the product of a-r and b-r.

The complete formula is therefore:

ab = r(a+b-r) + (a-r)(b-r)

How can we prove that this is correct? Let's work on the right side of the equation. First, look at


Using the findings form the previous section, this is the same as saying

ra + rb - rr

You can test it
with some numbers in place if you like:
e.g. 10(2+3+4) = 80
  and 20+30+40 = 80

Also – we met things like (a-r)(b-r) before when looking at fields of bananas.

To multiply (a-r) with (b-r), the pattern is “first two, last two, inside two, outside two”

Like this:

(a-r)(b-r) =

ab +rr -rb - ar

Putting the two halves together:

r(a+b-r) + (a-r)(b-r) =

ra + rb – rr + ab +rr -rb – ar

Notice that ra is the same as ar so we can write:

ra + rb – rr + ab +rr -rb – ra

We can also rearrange this to group terms (terms are the things in between plusses and minuses)

(ra – ra) + (rr – rr) +(rb – rb) + ab

No matter what values r and a have, (ra – ra) is zero, so we can drop that completely leaving:

(rr – rr) + (rb – rb) + ab

and we can similarly drop (rr – rr) and (rb – rb) leaving just ab.

This proves that our long equation is equivalent to multiplying a by b. But why would we replace a short problem like “a times b” with the long one:

ab = r(a+b-r) + (a-r)(b-r) ?

It is because of the relative sizes of the values, and the ease of multiplying r. If we choose r so that it is easy to multiply, like 10, 2 , 50, 100, 1000, 500 etc, and if our digits a and b are close to the reference number, then the calculations in ab = r(a+b-r) + (a-r)(b-r) are dead easy.

Here is another example:

if a=21 and b=19 what is ab?

Choose r=20

ab = r(a+b-r) + (a-r)(b-r) =

20(21+19-20)+(21-20)(19-20) =

20(21-1)+(1)(-1) =

20(20)-1 =

2(200)-1 =



You can do that in your head when you get used to the method.




Try 992 * 980

choose r=1000

ab=r(a+b-r) + (a-r)(b-r) =

1000(992+980-1000) + (992-1000)(980-1000)

1000(992-20) + (8)(20) =

1000(972)+160 =

972000 + 160 =


Check: ((992))((980)) = ((2))((8))=((16))=((7))


This method is actually a special case of a more general method. The more general method uses two reference numbers but in the case where the two reference numbers are the same, the calculating is shorter and easier. That special case is what we just looked at. I'll leave the more general case that uses two reference numbers for later so that we can do some fun things.

There are easier ways to add digits!

This is a method that I thought up while on the train. I'd be very surprised of no-one else has also worked out the same method.

Take the digits 2 4 6 8 and add them. In this example, you should get 20 fairly quickly by normal means.






halve each even number:





These are easier to add. They add to 10. Now double the 10 to get 20.

That was a very simple example to get you started. Let's lay it out like this so that we can progress with less page turning:

2	1
4	2
6	3
8	4
=	10

Will it work when there are some odd digits too?

9	?
1	?
8	4
4	2
2	1

What do we put where there is a question mark? We can add the two odd digits, and always get an even one, halve the result. In this case, 9+1=10 and 10/2 = 5, so we write one 5 in place of the 9 and the 1. Alternatively we could steal 1 from the 9, leaving 8 and give it to the 1 to get 2, therefore we can write 8/2=4 and 2/2=1

9	4
1	1
8	4
4	2
2	1

If adding 4 1 4 2 1 is still a challenge, then do it again:

9	4	2
1	1	
8	4	2
4	2	1
2	1	1
	=	6
=	12

The answer of 24 comes from 6*2=12 and 12*2=24

Don't get too excited yet, there is one more thing to learn. What do we do if there is an odd one left over as in this example:

8	4
4	2
2	1

There is not another odd number to share with the 9. What we do is imagine the 9 is an eight, and leave a clue for later by putting a tick mark near the 9.

9'	4
8	4
4	2
2	1
=	11

We double 11 to get 22, then add one for the trick with the 9. There will only ever be one or no ticks in any column of numbers. You just need to add one when you have doubled into a column that contains a tick. Here is a bigger example. See if you can follow it:

9	4	2
8	4	2
6	3	1
4	2	1
3	2	1
2	1	1
6	3	1
5	2	1
3	2	1
6	3	2
7	3	1
8	4	2
5	3	2
3'	1	1
2	1
	=	19
=	38




When you do this on paper, you can draw lines to link pairs of odd numbers. It makes it easier to see what's going on. You also have a couple of alternatives. Any two of the same numbers – odd OR even can be replaced by one of them only – as in the example above where two 1s in the middle column make only a single 1 in the far right column. You can also add a pair of odd digits and halve the result. Sometimes it's easier to do that– especially with the smaller numbers.

Another way to add a column of digits is to find groups of digits that add up to ten.


Here are the pairs: (4+6)+(8+2)+(9+1)+(4+2+4) = 40

and this is what is left over: 2+3 = 5

40 + 5 = 45

Check: ((24681294324))=((242424))=((11222424))=((12222))=((9))=((0))

and ((45))=((0))

Sometimes this works very well, but sometimes there are too many digits that are greater than 5, and there is not as many to add to ten. There is no reason why we can't combine the two methods:

9	8	4
8	4	2
  6 (cross this out ) 
  4 (cross this out )
7	8	4
  4 (cross this out )
  6 (cross this out )
7	8	4
8	4	2
	=	16
=	32

answer = 84 = ((3)) check:


Counting 11s

Trachtenberg describes another system. With this system, you count digits till you get to 11, then make a tick and start again. The leftovers and the ticks are combined in a special way. It's best illustrated with an example:

1 3 5 3
5 6 7'3
3'2'5 4'
2 3 4 8
7 6 2'1
7 9 1 8 Leftovers
1 1 2 1 Ticks

To get the answer, apply the weird rule of adding the columns, plus the right neighbour in the ticks row.

8 Leftovers + 1 Tick + no neighbour = 9

7 9 1 8 Leftovers
1 1 2 1 Ticks

1 Leftover + 2 Ticks + 1 Tick = 4

7 9 1 8 Leftovers
1 1 2 1 Ticks
    4 9

and so on, using a carry when the result is greater than 9. Be careful to include the leftmost invisible zero column. It's easy to forget it.

1 7 9 1 8 Leftovers
0 1 1 2 1 Ticks
2 0 2 4 9

Check: ((13535673325423487621))=((17))=((8) and ((20249)) = ((224))=((8))

There is a small trap. When the column leftover is 10, what do you do?

  1 4
  5 6	6 + 4 = 10
  2 3	Here, 6+3+2 = 11 so the 2 gets a tick
  7 6
1 7 0	There is no room for 10 in one column.
0 1 1	So you place a carry in the next Leftovers
  ---	column.
2 0 1	

Check: ((1456322376))=((12))=((3)) and ((201))=((3))

How to scam people using your mathematical tricks.

If you are an adult of legal drinking age, then one drunken night out with your mates down the pub, bet them some dollars that you can give them the cube root of any two digit number between 0 and 100 within a few seconds. You could ham it up a little and pretend to be more drunk than you really are. Oh – and don't try it when you are really drunk. This might decrease the odds in your favour. An example of course follows now.

“Bill. I betchew if you chake a chew digit munber and times it by itself then times it again and tell me whats the answer, ven I can chell you wo'tha' munber is even if you downtellme whatitwas. Heres a crackl-ator. I can only do it when I's drunk.”

Bill takes the bait and punches in some numbers and says, “Ok, fifty bucks to you if you do it and you pay me one hundred if you can't. You got 60 seconds mate. 405224”

Instantly, you know that the last digit of his number is 4. Then you take 405 and compare it against some secret knowledge and say, “74”. Bill looks startled, and wants to keep his $50, so he challenges you again, this time making it harder, “Crikey! That must be a lucky guess. You want to try that again? Double or nothing?”

“Sherrr. Hitme wivit.” you say.

Bill wan's to make it harder by choosing a bigger number so he says, “778688.” But you instantly say “92 mate. 'owbout a payout now?”

Bill has no chance of course, because you can instantly calculate every cube root that came from any positive two digit number. It takes some preparation and practice, but once you learn a few numbers of by heart, then it is very easy. Here is the method:

The second digit comes from this table:

X    L
1    1
2    8
3    7
4    4
5    5
6    6
7    3
8    2
9    9

X is the last digit of the cubed number. So if the cubed number is 175616, then X=6 and the last digit of the number that was cubed is L from the table above which equals 6. To get the first

digit of the number that was cubed, look it up in this table:

Y    R
1    1
8    2
27   3
64   4
125  5
216  6
343  7
512  8
729  9

For the number 175616, count three digits from the right and draw a line: 175 | 616. Throw away the digits on the right. 175 is greater than 125, but less than 216, so you choose R=5 and the answer is 56

You can do a mental check: ((175616))=((26))=((8)) and ((56))3 = ((2))((2))((2)) = ((8)).

In case you need a leg up with this mental check,

for 1+7+5+6+1+6 think:

2 times 6 = 12, 5 and 1 is 6, 3*6=18,

7 is one more than 6, 4*6=24 plus 1 + 1 =26. and 2+6 = 8 so the checksum is ((8)).

((56))3 = ((5+6)) times itself and itself again. But ((5+6)) = ((11)) which is ((2)).

((2)) times itself and itself again is ((8)).

So all you need to do is commit those two tables to memory. This is fairly easy. You only really need to remember the second table because the first one is contained inside it. If you look at the last digit of Y, you will notice that it is the same as L. Great! That's cut the work down. But it's easy to remember the first table anyway because 1 is 1, 9 is 9. 4, 5 and 6 is 4, 5 and 6. The rest are what we call 10s complement. This means that the number is what you would need to add in order to get to 10, so the 10s compliment of 7 is 3.

Also, the first table is easily calculated. It's the last digit of the cube. For example, 6 :

6 * 6 = 36 (ignore the three leaving 6) then 6 * 6 again is 36 (ignore the three) => 6

try with 7

7 * 7 = 49 (ignore the 4) 9 * 7 = 63 (ignore the 6) => 3

The second table is Y = R3 for R = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

Here is the working out for you to see:

1 x 1 x 1 = 1
2 x 2 x 2 = 8
3 x 3 x 3 = 27
4 x 4 x 4 = 16 x 2 x 2 = 32 x 2   =       64
5 x 5 x 5 = 25 x 5   =     125
6 x 6 x 6 = 36 x 3 x 2 = (72+36) x 2  = 108 x 2  =     216
7 x 7 x 7 =
8 x 8 x 8 = 2*2*2*2*2*2*2*2   =     512
9 x 9 x 9 =

I've missed out the workings for 7 x 7 x 7 because it needs more room for explanation.

7 x 7 = 49.

What is 49 x 7?

well, 50 x 7 is close, and it is the same as half of 100 x 7, and we can do that one.

100 x 7 = 700, so 50 x 700 = 350. But we only wanted 49 x 7, so 350 is 7 too much and the answer is 343.

For 9 x 9 x 9, it is the same as 81 * 9.

9 * 1 is 9 so we know it ends in 9.

9 * 8 is the same as 2 times 9 * 4 which is 2 times 2 times 9 * 2 = 2(2(18) = 2*36= 72.

Therefore 9 x 9 x 9 = 729.

It's only 7 x 7 x 7 and 9 x 9 x 9 that takes any work. We could just memorise those, but it's good to have a way to mentally compute the answer, especially if $100 is at stake.

You might want to memorise 6 x 6 x 6 as well, for speed.

I'll now go through another example, narrating my thoughts as I work it out.

Find the cube root of 328509.

Think: 328, hmm more than 5 cubed which is 125, what's 6 cubed? 6 is 3 x 2.

6 sixes are 36 and 36 doubled is 72.

72 plus another 36 is 108.

108 times 2 is 216

oh – so 328 is more than that.

What's 7 cubed?

7 * 7 = 49.

50 * 7 = 350

350 less 7 is 343

328 is less than that, so the first digit is 6, and the last one is 9 (from the first table).

Right. Having presented all that in the way that it was described when I first learned it, I can reliably say, “There is another way”. I've refined the trick so that you don't need to work out as much.

Here is the method. The unit's digit is easy to find by remembering the table. But if the table is too hard to remember, you can work it out easily. Take the number “157464” as an example. We know it is the cube-root of “something”. It ends in a 4. All you need to know is that the last digit is either a 4 or the 10 compliment which is 6. These are the only two options so we can quickly try one. If it works, then go no further and if not, then the other option is correct. Pick 4. Something-4 times Something-4 ends in 6 because 4x4=16. Then since 6x4 ends in 4 the cube must also end in 4... and it does. So “4” is the units digit.

For the tens digit, we can take advantage of two facts. The first is that there are only two digits in the number that we seek. The second is that it's checksum will tell us what it is if we can guess it closely enough. So for the number “157464”, we look at the three digits, “157” and guess if it comes from a tens digit if 1,2,3,4,5,6,7,8 or 9.

Clearly, it's not 1,2,3, or 4 because we know that 4x4x4 = 16x4 which is 64. So it seems a good guess to try 5.

Using “casting out nines” we try ((54))((54))((54)) = ((0))

(That's an easy one because 5+4=9 which gives ((0)) straight away.

This means that the number given must also have a checksum of ((0)).

So we try it:


There are other choices that could give is a checksum of ((0)), but they are poor estimates for the cube-root, and we throw them away on grounds of insanity.

To aid you further in the understanding and also to aid the visually dominant readers, I've prepared this graph that charts x-cubed for x in the range 1 to 9. You can see how it grows ever more rapidly, and can see that anything around the 500 to 700 mark comes from 8 or 9, and anything around the 200 to 400 mark comes from 6 or 7. The smaller ones are easy to calculate anyway.

x cubed
x cubed

Using this graph as an estimator, try finding the cube root of 226981.

Tens digit: looks close to 6 for the 226 part of 226981.

Units digit: Easy. 1x1x1 = 1

Try 61.

((61))3 = ((7))((7))((7))=((49))((7))=((4))((7))=((28))=((10))=((1))

and ((226981))=((226))=((10))=((1))

There are harder ways to earn $50!

Actually, to a mathematician, it looks a bit odd to put an exclamation mark after that $50. This is because the exclamation mark means something special.

1! means 1 you say, one factorial.

2! means 2 x 1 you say, two factorial.

3! means 3 x 2 x 1 you say, three factorial.

4! means 4 x 3 x 2 x 1 you say, four factorial.

so $4! is 4 x 3 x 2 x 1 = $24.

What is $50! ?

Here is a clue. According to my computer's calculator, 49X48x47x46x45x44x43x42x41x40x39x38x37x36x35x34 =


So imagine how much money is $50!

How much money is $50 factorial? And how big is big?

If my computer thinks 49X48x47x46x45x44x43x42x41x40x39x38x37x36x35x34 =


then how much money is $50! ?

My computer probably would not calculate it correctly. In fact to actually calculate the exact digit for $50!, you would have considerable trouble. However, there is a way to get an idea. It's an estimate called “Sterling's formula.” It uses “powers”. To raise something to the power 4, you take a number, say 7 and multiply it by itself four times. 7x7x7x7.

Here, we use it (and a calculator) to work it out roughly. When I say roughly – I mean you can trust the first digit or so – after that, it's just decoration because I have not analysed the rounding errors and method and can't state how accurate is the answer.

50 divided by 2.71828 = 18.3939844313

18.3939844313 raised to the power of 50 is


Then we compute:

2 x 3.1415927x50 = 314.1592700

and find the square root. That is to say the number that comes to

314.1592700 when you multiply it by itself.

That number is 17.7245386399.

Then we multiply

17.7245386399 by 1713131607977128345167867282703345846603127058970152073995279467.5006824621

which is about 30364467380824630429369408104150470387130303716885706336308454891.

That's an unimaginable amount of money. It's more than enough for any purpose whatsoever. In fact, it would be difficult to find a place to put it.

Another way to represent such a big number is to use some more notation to indicate how many zeros follow the number.

This is literally x10n where you put a digit in place of the little n to indicate the number of zeros. It makes sense because x102 means 10x10 = 100 and there are two zeros in 100. Similarly, x105 is 100000

We can say then, that $50! is very roughly about $3x1064 .

But how much is that? It's big money, but how big is big? Where would we put money like this? Will it all fit in a bank?

Let's try something. Let's compress each dollar into a cubic millimetre. There are a thousand millimetres in a meter, therefore there are 1000x1000x1000 cubic millimetres in a cubic meter. That's 10 and 9 zeros. We need 64 zeros. There are 1000 meters in a kilometre, so there are 10x1012 cubic millimetres in a cubic kilometre. Damn. Nowhere near big enough. It's about a thousand kilometres from Brisbane to Sydney. If there were a cube a thousand kilometres wide, then there are 1015 cubic millimetres in that. Phew! Already we have a structure that you could see from the moon, and it's nowhere near large enough to store $50! !

(Sorry – I could not resist putting the exclamation mark there. I hope you don't think I mean 50 factorial dollars factorial !)

Roll up! Roll up! Witness the magic of numbers!

This trick was invented by the physicist Martin Kruskal. He presented it as a card trick, but we can illustrate the principle with a string of random numbers.

3 4 3 6 1 6 7 9 8 7 2 7 8 1 4 2 3 6 6 4 2 9 8 9 2 5 6 2 3 8 7 3 9 1 5 9 8 3 7 8 6 8 2 7 4 1 4 4 3 9 7 8 4 1 4 9 9 5 7 9 6 1 1 7 5 4 2 6

Think of a number between 1 and 9. This is your secret number. Find the number at that position from the beginning. Let's say you pick 4, then your secret number is 6 because 6 is where you land after counting 4 digits. Your secret number is now 6. Count 6, and your secret number is now 7. Count seven and your secret number is now 3. Keep doing this until there are not enough numbers left. For example, if you land on the last 5 in the whole thing, then that is your final secret number because jumping 5 puts you off the end.

Right. Pick a number and try it.

Your last secret number is: 4

How could I know that? No, the list of numbers is not fixed. It works on playing cards too. You can shuffle a deck, assign 1 to the ace, and 5 to all the picture cards. You can ask someone to think of a number from 1 to 10, and keep it secret. Get them to deal the cards one by one and when they run out, tell them what their last secret number is. It works 5 out of 6 times. Those are good enough odds to make some people think that you are a mind reader.

But how do YOU do it? The same way that they do. Pick a number as a starting point and go for it. You might as well pick the first one since that increases your odds just a little.

Here is why it works.

Let's say you pick 4 as the starting point. I will make all the digits that you visit bold so that you can see how it progresses.

3 4 3 6 1 6 7 9 8 7 2 7 8 1 4 2 3 6 6 4 2 9 8 9 2 5 6 2 3 8 7 3 9 1 5 9 8 3 7 8 6 8 2 7 4 1 4 4 3 9 7 8 4 1 4 9 9 5 7 9 6 1 1 7 5 4 2 6

Now pick the first 3 as a starting point. I will make all the numbers visited an underline.

3 4 3 6 1 6 7 9 8 7 2 7 8 1 4 2 3 6 6 4 2 9 8 9 2 5 6 2 3 8 7 3 91 5 9 8 3 7 8 6 82 7 4 1 4 4 3 9 7 8 4 1 4 9 9 5 79 6 1 1 7 5 4 2 6

Wow! After only 6 hops, we land on a bold 9. From then on, it's lock-step to the final secret 4.

Try the second 3 as a starting point. I'll make these ones bigger.

3 4 3 6 1 6 7 9 8 7 2 7 8 1 4 2 3 6 6 4 2 9 8 9 2 5 6 2 3 8 7 3 91 5 9 8 3 7 8 6 82 7 4 1 4 4 3 9 7 8 4 1 4 9 9 5 79 6 1 1 7 5 4 2 6

I stopped after only 1 hop because it landed on a bold 6. From then on it's lock-step to the 4 again.

See if you can find a starting value in the first nine that don't land on the 4. By testing the secquences like this, you soon notice that it's highly likely that you land on a number that get's landed on from some other starting point. 

Casting out elevens.

My word processor has a word completion feature. It tries to guess the word that you need and offers it as an option. It's quite good, but when I typed “Casting out elevens”, it offered elephant instead of eleven. If you can cast out elephants, then all well and good but I can't. Anyway, casting out elevens is more complex than casting out nines, but it can detect more kinds of errors. Here is how it works. Let's cast out elevens for 87632487.

Starting at the right, the 7 is considered in an odd position. The 8 to its left is in an even position, the 4 is in an odd position, and the 2 is in an even position, and so on. We need to separate the odd positions from the even ones.

This is a good way:

7 2 4 7

8 6 2 8

Now you cast out 11s and put the total for each row to the right.

7 2 4 7 [[9]]

8 6 2 8 [[2]]

Notice that I decided on [[ ]] as a notation to indicated casting out 11s.

Then you say [[9]] – [[6]] to get [[3]] and that is the remainder.

Sometimes you will get a smaller number on the top. In that case, you can add 11 to it then do the subtraction. Here is one like that:

2 3 4 7 [[5]]

8 6 2 8 [[2]]

Since 5 is too small, you can add 11 to it. Why? Because we are casting out 11s, we are also free to put 11 in if it is convenient.

The problem is now:

2 3 4 7 [[16]]

8 6 2 8 [[2]]

The remainder is [[16]] - [[2]]= [[14]] = [[5]].

Lets try checking a multiplication problem.

Is 18456363 times 27 equal to 498321801 ?

[[ 18456363 ]]:

8 5 3 3 [[ 8 ]]

1 4 6 6 [[ 6 ]]-

=[[ 2 ]]

[[ 27 ]]:

7 [[ 7 ]]

2 [[ 2 ]]

=[[ 5 ]]

Then [[ 2 ]] [[ 5 ]] = [[ 10 ]] ( because 2 x 5 = 10 )

Now we check our answer:

4 8 2 8 1 [[ 12 ]]

9 3 1 0 [[ 2 ]] -

=[[ 10 ]]

That's great! It looks like we have a match.

You can cast out 9s as a double check.

((18456363 ))(( 27 )) = (( 0 ))(( 27 )) = (( 0 ))

and (( 27 )) = (( 0 ))

We are building up some good tools for checking our answers. One of the best tools is illustrated in the next section.

Baloney detection.

I borrowed this phrase form Carl Sagan's book entitled, 'The Demon Haunted World.” this book is a great read for budding scientists (and skeptics!). Part of the book tells you how to detect faulty logic and arguments.

Our baloney detection kit is a set of tools for estimating answers. Before diving into a problem, we ask, “What is a sensible answer?” and “What value am I expecting?” and “Are there two ways to do it?” and “Is there a simpler way?”

If we can estimate an answer – it's usually a simple method. It's a way of getting a rough answer quickly. And because it's quick and easy, it's more likely to be done correctly. Therefore when we do the exact method, our estimate acts as baloney-detection.

Let's try some examples.

102 times 76

Estimate: It's has to be a bit more than 7600 because 100 times 76 is 7600.

Calculation: 7590

Conclusion: There is something wrong.

102 divided by 76

Estimate: The answer is one and a bit because you can only fit one 75 into 100.

Calculation: 1.34

Conclusion: It looks about right.

104 + 345 + 873 + 124 + 234 + 756 + 234 + 5 + 34 + 92 + 232

Estimate: ( 1 + 3 + 9 + 1 + 2 + 8 + 2 + 0 + 0 + 1 + 2 ) * 100 = 2900

Calculation: 3033

Conclusion: It could be right.

Closer estimate: ( 10 + 35 + 87 + 12 + 23 + 76 + 23 + 10 + 3 + 9 + 23 ) * 10 = 3110

Conclusion: It's pretty close.

Casting out nines:

((3033) = ((0))

(( 104 345 873 124 234 756 234 5 34 92 232 )) =

(( 5 3 0 7 0 0 0 5 7 2 7 )) =

(( 5 3 7 5 7 )) =

(( 5 1 2 7 1 4 7 )) =

(( 5 4 )) = (( 0 )) so it is very likely to be correct.

Just in case you did not follow the demonstration. Let's look at the last example. The 104 is close to 100, 345 is close to 300 and so on. But to make it easier, I divided each number by 100 as well. So we get a string of numbers: 1 3 9 1 2 and so on. After adding those up, multiply the total by 100. It's just a little easier to add up the single digits than numbers in the 100s. Since more than ½ of the numbers were rounded down, we expect the estimate to be a little low. This is an estimate on an estimate! 92 gets rounded up to 100 of course.

Rules for rounding:

These numbers get rounded down to 0:






These get rounded up:






So to round the number 22736 to the nearest 1000, we do this:

22736 rounds up to 22740 ( that's rounded to the nearest 10 )

22740 rounds down to 22700 ( that's rounded to the nearest 100 )

22700 rounds up to 23000 (that's rounded to the nearest 1000 )

If you were buying material for curtains, then it would be nice to estimate the cost before you go into the shop. This is how you could do it: Count the windows in your house. Let's say you have 10 windows. Pick the most common size, and guess it's width and height. It's probably about 1m high and 1.5m wide. Curtains are folded even when closed, so you need about twice the width of the window. This means that we can call each window 3m wide, and 1m high. So 3 square meters is required for each window. With 10 windows, you expect to buy about 30 square meters of material. Curtain material usually comes in a 1m wide roll that is folded. So the unfolded width is usually about 2m. You would need 15 meters at 2m wide. How much would this cost? You probably have no idea, but if you have a budget of $1000, then it would be nice to get 15m of material for under $1000. Estimate how much is 1000 divided by 15.

15*6 = 90 and that's nearly 100.

1000*6 = 6000

Our estimate is $6000 divided by 100 which is $60. This means that we can afford to buy material that costs less than $60 a meter. I expect in today's prices, that's a decent quality material, and may even include the cost of backing and little fittings.

(The actual calculation of 1000 divided by 15 is 66 and two thirds – but why bother calculating it exactly if we only guessed at the size of the windows in the first place?)

With this estimate, we can measure each window and do a proper calculation. The final result has to be reasonably close to our estimate, otherwise we say, “Baloney” and check the measurements.

Does Algebra scare you?

Algebra scares many people. “You can't add letters! Don't be silly!” But algebra is really not that bad. The rules are mostly easier than those for many electronic games. Sometime in high school you will meet the formula for a quadratic equation. Here is is. Avert your eyes!

Many children say, “Why am I learning this? I will never use it. It's a load of rubbish.” Perhaps. Perhaps not. But the same can be said of many things learned at school, and the same can be said of many things learned in a sports-meet. In fact, most people will not become a professional sports person. So why bother learning the rules of rugby or cricket? Of course we all know the answer to that one. It's good for the whole person to do sport, and some will become professional cricketers or footballers. The same can be said for the above equation. Understanding it is good for the whole development of the mind, and some students will become engineers.

Some of the problems with the scary-looking thing above is that it's never introduced. By this I mean, teachers don't say, “Hey Bill, I'd like you to meet quadratic. He's a friendly chap really. A bit rough around the edges, but once you get to know him, he's just a puppy dog at heart. Quadratic is really good for Engineers, and architects, and sending people into space. You could meet him many times in the future.”

I won't be delving into the details here. (Yes, you can relax now) If I did, then it would take many pages, and would not fit in with the flavour of the rest of the text. So you might be thinking, “Why tell it then?” The reason is that I want to show that there is another way to look at it... and it's not that scary.

Take this sum: 5 + 4

We can draw this like so:

We feed “5” and “4” into the “+” and the answer pops out the bottom.

If either the 5 or the 4, (or both) are changed to a word, like “left value” or “right value”, then you are instantly doing algebra. Like this: Change the left value to a 6. Now the answer is 10. Of course, mathematicians are lazy, so it's cheaper to say x instead of “left value”. You can pick any word or letter, or shape to represent a number, but since there are all kinds of numbers, like counting numbers, and decimals, and fractions, and even something called “imaginary” numbers, then mathematicians agree to use a familiar symbol in many cases. Its' really handy that this notation is quite the same all over the world. So mathematical formula look the same in any country. The bad thing is that may people get scared of the mass of unfamiliar squiggles and weird notation, and it can be hard to find explanations. Doing mathematics is a little like Karate. You have to start at the white belt, then build up what you learn gradually until you become a black belt. (Then the process starts again.)

We are doing white-belt.

This diagram-notation can be used to represent the quadratic:

(It says, “minus b plus or minus the square root of b squared minus four a,c all over 2a)

Here it is:

The quadratic formula
The quadratic formula

Now it looks more like a game. The blue circles are called parameters. The letters represent numbers that come from a certain place. The yellow circles are called operators. They operate on the parameters. I drew the arrows to link parameters that are the same.

If someone said a was 2, and b is 11 and c is 5 , you could probably get two answers popping

out the bottom without me explaining how to do it. But just in case you want a leg-up, this is how it works:

2a = 4

4ac = 40

b2 = 121

121 – 40 = 81

The square root of 81 is 9 

Now there are two paths. One leads to the answer that pops out the bottom left, and the other to the right. Let's do the one that leads to the bottom left.

-b = -11

-11 + -9 = -20

2a = 4 (we already know that)

-20 / 4 is our first answer.

= -5

The other one that leads to the right:

-11 +9 = -2

2a = 4 (as before – we just copy that one down)

-2/4 is our second answer.

= -0.5

It's not so hard. Doing it this way, with the diagram makes it less scary, and makes the formula more familiar.

See if you can work out how the quadratic as an algebraic equation is equivalent to the diagram. Before you rush off trying to convert all your formulas to pictures, note that some operators can have their parameters fed into them in any order, and the answer is the same in each case. This is true for times, and for plus, but not for divide and for minus. So a diagram like this would also need rules so that the reader knows which parameter to feed in before the other. You could agree at the start for instance that things higher up the page go in first. In this way, for example 5 - 4 = 1 is a different arrangement compared to 4 – 5 = - 1

Does my diagram follow rules? What is the rule for operator order? Study it. Perhaps you can improve it.

Junk food.

I used to have a bit of trouble with takeaways at school; not the junk food, but the thing where you take one number away from the other. So here are a two alternative ways to do take away.

The way that I was taught:

 49 -

8 take away 9. You can't. Borrow one from the column to the left. Reduce that by one and add ten to the 8.

6 618	
  4 9 -

Now you can say 18 – 9 = 9 and 6 – 4 = 2 and the 6 is left alone.

6 618
  4 9 -
6 2 9

Of course it works. But there are two drawbacks. Crossing out makes it messy. 18 – 9 is a two digit number less another. We can make it simpler.

As a child, I noticed that the number below is the 10s compliment of the bottom one minus the top one. So it was easier for me to look at the 8 and the 9, see that 8 is too small, then do 9 – 8 = 1. For some reason, I just see that 9 is the number that you add to one to get 10. Effectively, doing 10 – 1 was easier for me than doing 18 -9 and all that crossing-out.

 49 -

The penalty for this trick is that we still need to borrow one from the top number in the next column. But I thought it was easier to add one to the bottom number in the next column instead. And it is easier. Adding one involves a little tick mark, while taking one away means crossing out and re-writing. Mess always lead to errors for me.

 6 7 8
   4'9 -

Then it's 7 – (4+1) = 2. Do the 6 and that's it.

 6 7 8
   4'9 -
 6 2 9

 Now the good thing about this method is that you can do all the ticks first, then do the taking away on each digit, in any order.

Step 1. Identify all the columns where the top number is too small.

    | |   | |   |   | |
6 8 7 7 9 4 5 7 3 8 1 1
  4 9 9 8 7 6 7 8 5 9 9 -

Step 2: For each of these columns, you put a small tick mark by the bottom number in the column just to the left of those you marked where the top number is too small. This is to remind us that we need to take one more away in that column.

6  8  7  7  9  4  5  7  3  8  1  1
   4' 9' 9  8' 7' 6  7' 8  5' 9' 9 -

Step 3: Having done that, scan each column again, noting that the tick mark makes its number one greater. You are looking for new columns where the top number is too small. In this case, the 5th column from the right has 7 above a 7 that has a tick mark, so this really means 7 less 8. so 7 is now too small, and you put a tick mark in its left column as well.

6  8  7  7  9  4  5  7  3  8  1  1
   4' 9' 9  8' 7' 6' 7' 8  5' 9' 9 -

Step 4: In any order, you can calculate each column's answer. We will do the easy ones first.

6  8  7  7  9  4  5  7  3  8  1  1
   4' 9' 9  8' 7' 6' 7' 8  5' 9' 9 -
6  3        0              2

Step 5: For the remaining columns, we need our knowledge of 10s compliment. Each time there is a tick mark, as in Step 4, you think of that number as being one bigger. The tick mark is worth exactly one. The important thing here is you still do take-away, but this time in two stages. The first stage is take the bottom column away from the top, then you write down the 10s compliment. The tens complement is the number that you need to add to get a total of 10. It's very easy. The tens compliment of 1 is 9 because (1+9) = 10.

Let's do the first vacant column where we have 7 and 9'

9' is the same as 10

10 – 7 = 3

We need to add 7 to 3 to get 10, so the answer for that column is 7.

Here is the answer with checksums.

6  8  7  7  9  4  5  7  3  8  1  1     ((3))
   4' 9' 9  8' 7' 6' 7' 8  5' 9' 9 -   ((0))
6  3  7  8  0  6  8  9  5  2  1  2     ((3))

Try a few of these on your own. You will soon find that this method is cleaner, faster, and more reliable than the usual method.

What do you do when the top number in total is smaller than the lower number as in the case

234 – 987?

You know the answer is a negative number. The easiest way to compute this is to put the bigger number on top, do the subtraction as normal, but call the answer negative.


5 – 10

is the same as:

minus ( 10 – 5 )

= - 5


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    Post Comment

    • Manna in the wild profile imageAUTHOR

      Manna in the wild 

      6 years ago from Australia

      Thank you Gary and Mahesh.

      Mahesh - the link to your site is ^^ up there.

    • profile image


      6 years ago

      This one is nice. I came here to see if Mathematics helps in fostering intelligence and memory in children. I found your website. Your website is really good. Full marks to you for this.

      Can you please visit my website and give me your feedback. I have developed an arithmetic puzzle game based on the concepts of mathematics. Can you give a link to my site also.

    • Manna in the wild profile imageAUTHOR

      Manna in the wild 

      7 years ago from Australia

      Enjoy your beer! Thanks for the score.

    • Gary Shorthouse profile image

      Gary Shorthouse 

      7 years ago from Reading, UK

      Well, I will come back and finish it later. Halfway through and I feel a beer coming on.

      Actually scored it beautiful as that's what it is.

    • Manna in the wild profile imageAUTHOR

      Manna in the wild 

      7 years ago from Australia

      Thanks Cheeky. It's here for reference. Hopefully someone in "teaching" will find it useful.

    • Cheeky Girl profile image

      Cassandra Mantis 

      7 years ago from UK and Nerujenia

      Holy crap! This took a long time to read, and maybe that is why there are no comments on it. I would suggest maybe breaking up hubs like this into several parts and theming them somehow, but Congrats on a fine hub on Maths! You deserve an Oscar for writing all this out and putting so much effort into it! It is remarkably easy to follow! Cheers! Rating this up, you deserve it!


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