More division tricks. Numbers ending in 1 or 2

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#1 5 October, 2017 - 06:18
Joined: 6 years 5 months ago

More division tricks. Numbers ending in 1 or 2

The same trick we did with division by numbers ending in 8 or 9 can be done with numbers ending in 1 or 2.

However; instead of adding numbers to the remainder we now subtract numbers from it.
An example will make this clear.
We start with dividing 10 by 6:
10/6=1r4. We now subtract the digit found (1) from the remainder. 4 becomes 40, from which we subtract 1 to get 39.
We continue with the remainder of 39. 39/6=6r3. the remainder of 3 becomes 30 and from 30 we subtract 6 to get 24
24/6=3r6 => 57
57/6=9r3 => 21
21/6=3r3 => 27
27/6=4r3 => 26
26/6=4r2 => 16
16/6=2r4 => 38
38/6=6r2 => 14
14/6=2r2 => 18
18/6=2r6 => 58
58/6=9r4 => 31
31/6=5r1 => 05

Answer: 1.639344262295

Numbers ending in 2.
We now subtract twice the digit from the remainder.

10/6=1r4. 40-2=38
38/6=6r2. 20-12=8
8/6=1r2. 20-2=18
18/6=2r6. 60-4=56
56/6=9r2. 20-18=2
2/6=0r2. 20-2*0=20
20/6=3r2. 20-6=14
14/6=2r2. 20-4=16
16/6=2r4. 40-4=36
36/6=5r6. 60-10=50
50/6=8r2. 20-16=4
4/6=0r4. 40-0=40
40/6=6r4. 40-12=28
28/6=4r4. 40-8=32
32/6=5r2. 20-10=10
We started with 10 and we now have a remainder of 10, so the digit sequence repeats from the top.

Answer: 1.61290322580645161290322580645161...

9 October, 2017 - 13:26
Joined: 10 months 3 weeks ago


did you realize that this technique is the very same as described by greymatters as leapfrog division II? Here is his pst:

I however find your way of doing it a bit easier to manage, as it does not involve subtracting 1 from the dividend. On the other hand, there seem to be pitfalls:

In your first example you do

24/6=3r6 => 57

where I would say that

24/6 = 4r0

So what is the rule in such cases, why not divide the whole thing?

I tried the technique using this example: 46.20 € (for a monthly bus ticket) divided by 21 tours is how much money per ride? I can’t seem to find how to do it:

4620 / 21
4620 / 2 = 2310 r 0

11 October, 2017 - 02:37
Joined: 6 years 5 months ago


did you realize that this technique is the very same as described by greymatters as leapfrog division II?

No, I did not.
Interesting way of dividing though. I like it.


In your first example you do
24/6=3r6 => 57
where I would say that
24/6 = 4r0
So what is the rule in such cases, why not divide the whole thing?

Because in the next step we need to subtract. A remainder of zero leaves no room for subtraction of course.

Let's do 4620 / 21

(Of course, 4620 is 4200 + 420, so division by 21 is easy in this case.)
However, let's use this technique.
4/2=2r0. 6-2=4.
4/2=2r0. 2-2=0.

Answer: 220.

In my mind, I always use the apples and baskets analogy when doing these kinds of shortcuts.
It helps me to regenerate these algorithms if I am unsure whether to for example first subtract and then shift the decimal separator or the other way round.

In this case, if you work out 4620/21, you would normally start with 46/21 = 2r4.
So when using a shortcut you need to come to the same result.

In this case we do two things to make the division simpler.
First, instead of dividing 46 by 21 we divide by 20 (and later make the correction )
Second we realize that 46/20 gives the same integer part as 40/20. So for the digit in the answer, we might as well use 40/20 = 4/2 and then for calculating the remainder we take the 6 into account.

Going back to apples and baskets:
I have 46 appels that need to be put into 21 baskets. I first put 2 apples into 20 baskets, leaving 6 remaining. Then I add 2 into the 21st basket so now I have 4 remaining. This is the subtraction part.

This is exactly what we do when we started the calculation with::
4/2=2r0. 6-2=4.

This is why we add to the remainder in

In the example of 100/59, we start with 10/6. In terms of apples and baskets we put 1 apple into 60 baskets, leaving 40 remaining. We only need 59 baskets full, so we take the apple out of the 60th basket and add it to the remaining 40 to get 41 remaining apples.

11 October, 2017 - 02:40
Joined: 6 years 5 months ago

Btw. I am working on a post for division by 23 using both moving up to 25 and moving down to 20 and looking at the difference.
Should be done in the next couple of days or quicker if I find time to write.

16 October, 2017 - 07:26
Joined: 6 years 5 months ago

TorstenBerg wrote:

did you realize that this technique is the very same as described by greymatters as leapfrog division II? Here is his pst:

Let's talk more about the leapfrog division.
Because upon closer inspection, you are so right!

At first I could not understand why we need to subtract one from the dividend and then subtract every digit found from nine.

But now I understand it better.
And I assume it is not immediately obvious for everybody.
Btw., I find Greymatters algorithm very elegant, especially now I understand better why it works! So nothing I say should be thought of a criticism.

This is what the algorithm does and is not explained:

In the example of 13/21, 13 gets changed to 12.999999...., which is allowed of course.
This eliminates the need to look ahead like I need to do. So in that sense it is better than what I do.
In the calculation of 1/11, 1/21, 1/31, 1/41, etc. the answer is a repeating string of digits.
And because of that changing 13 into 12.99999... does not give any problems.
The nines of course make subtracting easier.
That is the reason '13' needs to change to '12', and that is also the reason we need to subtract from nine every time.

Keep in mind though that this algorithm works only for expanding fractions.
In the case of 6003/21, one needs to first split the integer part and the fractional part (to get 285 18/21). Mine is more general, since I use it for both.

A funny thing happens if you want to expand it to fractions ending in 2. Then we don't subtract the digit from nine, but then we need to subtract twice the digit from nine.
If we find digits bigger than 4, then subtracting twice from nine gives problems of course.

If we do this for 32, we see that we need a special way of looking ahead in order to stop. 10/32 = 0.3125. So this is not a continuous stream of digits. Now we need a way to find out that we are done.
Again; with fractions of 31 we subtract each time the digit from nine. Now we need to subtract twice the digit from nine.

9/3=3r0. (9-2*3=3)
03/3=1r0 (9-2*1=7)
07/3=2r1 (9-2*2=5)
15/3=5r0 (9-2*5=-1)
0[-1] ???????

We have found all digits, so the 0[-1] must indicate a stop, that we are done.
One would expect to stop at '00', i.e. when there is no remainder.
However, 0[-1] actually means 'no remainder'.
Since we are working with a continuous stream of nines, 00 is actually 0[0,99999999...]. We only concern us with the first 2 digits: 00.

We see this at 10/41.
Using the LeapFrog, we get:
1: 9/4 = 2r1.
2: 17/4=4r1
3: 15/4=3r3
4: 36/4=9r0
5: 00/4=0r0
6: 09/4 = 2r1
7: 17 => repeat from line 2 above
Answer: 0.2439 02439 02439 02439 ...

We see that at line 5 we need to continue with '00'.
And again, the reason is the line of continuous nines that we are omitting.
'00' is actually '00.999999'.

In short, we can use the 'Leapfrog Division II' only for fractions that end in 1 (11, 21, 31, etc.). For fractions like 12, 22, 32, 13, 23, 33, etc. it is less useful.

That does not make it bad in my opinion. It works great for fractions like 31 and it eliminates the need for looking ahead that I need to do.
And keep in mind that a lot of fractions can be changed into ending in 1:
1/17 = 3/51

Like I said, very elegant for fractions like 31.

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