Associated Topics || Dr. Math Home || Search Dr. Math

Arithmetic in Other Bases

```
Date: 02/25/2002 at 09:24:11
From: Linda
Subject: Arithmetic in other bases

It is easy to perform the four basic algorithms in base 10 form.  But
we have recently been asked in my math class to perform arithmetic in
other bases. For example: 5430/13 (base 6). What I need to know is how
to write out the steps for addition, subtraction, multiplication, and
division in other bases. Please help or let me know where I can go for
help.
```

```
Date: 02/27/2002 at 01:52:45
From: Doctor Mike
Subject: Re: Arithmetic in other bases

Hello Linda,

It's exactly the same, only different. If you follow the familiar
processes, but keep in mind the consequences of the digits meaning
powers of six instead of powers of ten, you will have it. Let's
use your example for starters.

Remember EVERYTHING is in base 6. To divide 13 into 5430 by the usual
long division, you first say that since 13 cannot go into 5, you need
to find out how many times 13 can go into 54. My first guess was
4 times. But, (in base 6) 4 times 13 is 100, which is too big, so it
must go 3 times. So, 3 is the first digit of the quotient, and what
you write under the 54 is 3 times 13, which is 43. This really looks
strange, because we are all VERY used to multiplication tables in
base 10, so facts like 3*13 = 43 or 5*5 = 41 take some getting used
to.

Getting back to the division, write 43 under the 54 and subtract to
get 11. Bring down the next digit from the dividend to make it 113.
If you remember the "false start" we made, namely, that 4*13 = 100, it
should be pretty easy to do in your head that 5*13 is 113. That means
that 5 is the next digit in the quotient, and 5*13 = 113 is written
down below. Subtract 113 from 113 to get zero. Bringing down the final
digit 0 from the dividend shows that we are done, and the division is
exact, with quotient = 350 and no remainder.

I will try to do this semi-pictorally to give you an idea.

3 5 0
----------
1 3 / 5 4 3 0
4 3
-----
1 1 3
1 1 3
-------
0 0

One more example: Add 543 and 321. First add the 3 and the 1 in the
units place to get 4. Write that in the units place in the answer.
Next, add the 4 and the 2 in the six's place to get 10, so write 0 in
the six's place in the answer, and "carry" the 1 to the six squared
place. Add the 5 and the 3 in the six squared place to get 12, and
then add on the carry to get 13, and write that all down below to get
the full answer 1304.  See?

Visually,

1
5 4 3
+ 3 2 1
-------
1 3 0 4

I hope this helps. To be honest with you, even as a mathematician
I do NOT have the base 6 multiplication tables memorized. To do the
5 squared calculation, I first remembered the base 10 fact that
5 squared is 25 in base ten, and then figured out that in base 10 we
have that 25 = 4*6 + 1 so that 25 base 10 equals 41 base 6. See?
As easy as pi, right?

- Doctor Mike, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 02/27/2002 at 08:06:49
From: Linda
Subject: Arithmetic in other bases

Thank you very much for your help. You have helped out a lot!
```
Associated Topics:
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/