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### Long Division in Base 36

```Date: 11/05/2003 at 07:12:21
From: Nick
Subject: Divsion in a base 36 system

I am trying to use an "imaginary" system that is base 36 (ie. 1, 2, 3,
4, 5, 6, 7, 8, 9, A=10, B=11, ..., Z=35) and I am struggling to divide
one number in this system by another using long division.  I assume it
would be similar to division in the hexadecimal system but I would be
grateful for any help you could give me.

Cheers, Nick

```

```
Date: 11/05/2003 at 12:56:58
From: Doctor Peterson
Subject: Re: Divsion in a base 36 system

Hi, Nick.

Have you seen our FAQ on bases, which includes links to answers on
division?

Number bases
http://mathforum.org/dr.math/faq/faq.bases.html

Division in any base uses exactly the same method; you just need a
different multiplication table.  Base 36 is particularly awkward,
since its table is so large; I would not actually write out a table
(unless I had a sadistic teacher who made me do a lot of these!), but
would just make a table to allow me to quickly convert digits to base
ten, and then do the actual arithmetic in base ten where necessary.

So let's do it.  Here are the digits:

0=0    A=10    K=20    U=30
1=1    B=11    L=21    V=31
2=2    C=12    M=22    W=32
3=3    D=13    N=23    X=33
4=4    E=14    O=24    Y=34
5=5    F=15    P=25    Z=35
6=6    G=16    Q=26
7=7    H=17    R=27
8=8    I=18    S=28
9=9    J=19    T=29

Now let's first do a multiplication for practice; I'll multiply 4P by
G3:

4P
* G3
----

Remember that

4P = 4*36 + 25 = 169
G3 = 16*36 + 3 = 579

So the answer should be

169 * 579 = 97851

which we can convert back to base 36 by dividing by 36 repeatedly:

97851 / 36 = 2718 rem 3
2718 / 36 =   75 rem 18 = I
75 / 36 =    2 rem 3
2 / 36 =    0 rem 2

So our answer should come out to 23I3 (base 36). Let's see if we can
do it.

We'll have to multiply four pairs of digits; let's translate each of
those pairs into decimal, multiply, and convert back:

3*P = 3*25 = 75 = 23
3*4 = 12 = C
G*P = 16*25 = 400 = B4
G*4 = 16*4 = 64 = 1S

Now we use the usual algorithm to multiply:

B
2
4P
* G3
----
E3 <- 3*4P
234  <- G*4P
----
23I3

Good. We can multiply successfully!

Now let's reverse that and divide 23I3 by 4P. We should get G3.
_______
4P ) 23I3

We first have to estimate 23I/4P, which we can do by finding 23/4. In
decimal, that is 75/4, which is about 19, or J. Let's try J:

D
4P
*  J
----
2H7

That's bigger than 23I; the excess is E6, which is about 3 times 4P,
so let's subtract 3 from J and try G:

4P
*  G
----
234

That looks good; so we can use it:

____G__
4P ) 23I3
234
---
E3

Now we have to divide E3 by 4P; E/4 = 14/4 (base 10) is about 3, so we
try 3:

4P
*  3
----
E3

Ah! It's exactly right:

____G3_
4P ) 23I3
234
---
E3
E3
--
0

So the answer is G3 (which we knew all the time, but ignored).

Now, everything I did here is exactly what I would have done if I were
teaching you to divide in base ten, except that we had to convert
numbers to base 36 to make our guesses (which we could have avoided if
we had a whole multiplication table to work with), and had to multiply
and subtract in base 36 to get the actual answers. It's very simple,
but very ugly. And I'm doing it now so I can put it in our archive and
never have to do it again!

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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