 # Fast division by 23

In two threads I showed how to make a division simpler by changing the divisor to a round number that is easier to work with:

Now let’s take a division by 23. We can either move 23 up to 25, or we can change it down to 20. Let’s do both and see which one is easier.
Btw., I consider 25 a ‘round’ number in terms of division. ‘Round’ being easy to use. Division by 25 is the same as multiplication by 4.

First moving up to 25.
The difference 25-23 is 2 so we correct the remainder by a factor of 2:

100/23
100/25 = 4r0 0+24=8
80/25 =3r5. 5+2
3=11
110/25 = 4r10. 10+2*4=18
180/25 = 7r5. 5+14 = 19
190/25 = 8r-10. -10+16 = 6
(if you started with 190/25=7 you quickly would have realized that the remainder gets too big: 190/25=7r15. 15+14 = 29. 29 > 25, so you have to use 8 instead of 7. It might feel strange at first (to use negative remainders) when you start to calculate this way. One way of dealing with this is to look ahead how much you need to add to the remainder and seeing if that does not become too big.)
60/25 =2r10. 10+4=14
140/25 = 6r-10. -10+12=2 (again 140/25=5 gives a remainder that is too big. So we have to use 6 instead of 5).
20/25 = 0r20
200/25 = 8r0. 0+16=16
160/25= 6r10. 10+12 = 22
220/25 = 9r-5 -5+18=13
130/25 = 5r5. 5+10=15
150/25 = 6r0…

Compare this with rounding down.
We could round 23 down to 20 and do 100/20. Even better, we can also do 10/2 and correct 3 times the digit in the remainder.
Keep in mind that after we divide by 2, we need a remainder that is large enough to subtract 3 times the digit from.

10/2 = 4r2. 20-34 =8 (10/2 of course is 5r0. However, this does not leave a remainder to subtract from. Therefore we take 4r2 instead of 5r0)
8/2 = 3r2. 20-3
3=11. (8 divided by 2 usually is 4, but again this leaves no remainder)
11/2 = 4r3. 30-12=18
18/2 = 7r4. 40-21=19
19/2 = 8r3. 30-24=6
6/2 = 2r2. 20-6=14
14/2= 6r2. 20-18=2
2/2=0r2. 20
20/2 = 8r4. 40-24 = 16
16/2 = 6r4. 40-18=22
22/2 = 9r4. 40-27=13
13/2= 5r3. 30-15=15
15/2= 6r3…

With a bit of practice both ways can be executed quickly.
When rounding down, you need to ‘look ahead’ how much you need to subtract in order to have a remainder big enough.
When rounding up, you need to ‘look ahead’ how much you need to add in order to have a remainder small enough.
In the 18/2 part we just did you need to go down even two steps:
18/2 = 9r0. 0-39 is negative.
18/2 = 8r2. 20-3
8 is negative.
18/2 = 7r4. 40-3*7 is positive, so we need to use this one.

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That was another great algorithm Kinma. But besides such brute-force division, one could also instead memorize the cycle of 23. This has just 11 repeating digits, in particular: 04347826086 and then continue with their exact complements (to 9) respectively: 95652173913.

100/23 = 4 with a remainder of
8/23 = 0.347826086 95652173913 04347826086 95652173913 04347826086 and so on.

This trick is just recital with not much calculation needed, besides finding the initial remainder and the 1st decimal digit’s position inside that sequence. After that, it only requires memorizing just 11 sequential digits 04347826086 and then continue by subtracting their complements to get 95652173913 , hence you can divide by 23 as fast as you can recall your recital. Albeit not mental calculation, I consider that to be faster than brute-force division. I guess Daniel Tammet thought the same when he was dividing with prime numbers like 97 using his preset memorization of 48 digits.

And since this is a primarily memory forum, I think it’s useful to incorporate some extra memory tricks into calculation.

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Alternative method for dividing by 23: Multiply the dividend by 3, and then divide it by 69, using the technique for dividing by numbers ending in 9.

So, 100/23 becomes 300/69, and you still get the same answer.

So true.

This is true and can do with a bit of repeating.
Like here:

Excellent find! That works just as well of course.
One of the things I like in mental calculation is that there are usually even simpler ways of calculating.
Creativity is so important!

One useful thing in memorizing full reptend primes (such as 23) is that they’re also sums of geometric series.

For example, 1/23, as has been mentioned, is equal to 04347826086 95652173913 repeating.

This is the sum of:
``` 0.04 0.0032 0.000256 0.00002048 0.0000016384 ```

…and so on. Written as a geometric series:

Σ 0.04*((0.08)^n) for n = 0 to infinity

…which is equal to:

(0.04)/(1-0.08) = 0.04/0.92 = 1/23

There’s also more than one way to express full reptend primes as a geometric series, as well. This also works starting with 0.043478 and repeatedly multiplying by 0.000006:

Σ 0.043478*((0.000006)^n) for n = 0 to infinity

…which is equal to:

(0.043478)/(1-0.000006) = 0.043478/0.999994 = 1/23

Despite this one being longer, I find it a little more memorable, as you can think of 043478 followed by that number multiplied by 6, which is 260868 (but remember that the 8 in the units place becomes a 9, making it 260869). That 9 at the end is your clue to start subtracting from 9s, as discussed by Nodas:
``` 043478 26086 956521 73913```

Sometimes seeing patterns in things can help simplify the memorization.

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Thank you, Grey Matters for - once again - a fantastic post!

I found another interesting pattern for 23rds, which will help you remember the digit sequence. It’s slightly different for the first half between the 0 and the first 9 than it is for the second half between the 9 and the repeating point.

```1/23=0. 0 43 47 82 60 86 9 56 52 17 39 13```

First half (starting at “43” and ending with “86”):
To the first pair, add an imaginary 1 at the beginning, and divide by 3 (ignoring any remainder). To the next pair, add an imaginary 2 at the beginning, and divide by 3. To the third pair, add an imaginary 1, and divide by 3. Keep repeating this pattern until you get to the 86.

So, 43 becomes 143, and 143/3 = 47 (the next pair)
47 becomes 247, and 247/3 = 82 (the next pair)
82 becomes 182, and 182/3 = 60 (the next pair)
60 becomes 260, and 260/3 = 86 (the final pair of the first half)

Second half (starting at “56” and ending with “13”):
To the first pair, add an imaginary 1 at the beginning, and divide by 3. To the next pair, don’t add any prefix to the beginning, and just divide by 3. To the third pair, add an imaginary 1 at the beginning, and divide by 3. Keep repeating this pattern until you get to the 13.

So, 56 becomes 156, and 156/3 = 52 (the next pair)
52 stays as 52, and 52/3 = 17 (the next pair)
17 becomes 117, and 117/3 = 39 (the next pair)
39 stays as 39, and 39/3 = 13 (the final pair of the second half)

Grey Matters, I love it. Great find!
Feels like magic.

I simply learn the sequence of 23.
I use major system to do this.
23 - becomes Name in major system
Repetition digit - Sir Mark Vangsvag
04347826086
And rest part is simple - 95652173913
( complement of first part)