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Octal Multiplication confusion

by mr_coffee
Tags: confusion, multiplication, octal
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Oct9-05, 11:11 AM
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I am reading this table in the bookk and everyhting makes senes but i'm confused on the last part, under the Octal. For example the first line:
5x2 = 10 = 8+2 = 12, i see where the 2 came from in 12, but where did the 1 come from...also
5x6 + 1 = 31 = 24 + 7 = 37, the 7 makes sense but where did the 3 come from? Thanks. Here is the table.
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Oct9-05, 11:47 AM
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Wow, that's bad notation. 10 is not equal to 12. 1010 = 12[/sub]8[/sub].

810 + 210 = (1 x 81) + (2 x 80) = 128

2410 + 710 = (3 x 81) + (7 x 80) = 378

Do you get it now? For comparison, consider:

1000 + 200 + 30 + 4 = (1 x 10³) + (2 x 10²) + (3 x 101) + (4 x 100) = 123410
Tom Mattson
Oct9-05, 11:54 AM
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Mr Coffee, please:

1. One thread per topic.

2. Homework goes in the Homework Forums.

I've deleted your other thread that is identical to this one, and moved this one out of the EE Forum.



Oct9-05, 12:31 PM
P: 1,629
Octal Multiplication confusion

Thanks akg! i get it now! sorry Tom, I saw other people posting homework questions in EE dealing with these type of problems, but i'll keep them in this forum from now on!
Kenneth Mann
Oct10-05, 02:22 AM
P: 410
What an abomination!

Just FYI: We learned the decimal multiplication tables in the second or third grade (at least, most of us did), so that we can multiply decimal numbers. We never learned Octal multiplication tables, so we multiply in octal by converting back and forth -forth and back. There's an easier way. Octal converts readily to binary, and binary multiplication tables are simple (0 X 0 = 0, 0 X 1 = 0, 1 X 0 = 0, and 1 X 1 = 1; that's all there is to it). Thus, to multiply these numbers we get:

_____ becomes

100,011,111,111,010 (the commas were used in the intermediate items just keep the spacing in the awful HTML world) Then, converting this back to octal, in our heads, we get:


A computer would do the multiplication in binary simply by --- shift and add --- shift and add --- etc.

Oct11-05, 02:35 AM
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Many years ago I designed supercomputers. It turns out that some of the Burroughs mainframes were built primarily as octal machines rather than binary machines.

This was primarily evident in their addressing. They had a 48-bit word which consisted of 16 octal digits. They could address by words, and they also could address the eight 6-bit bytes in a word with pointers similar to modern microcomputer byte oriented pointers.

This was a very convenient system.

Then one day they decided to convert it from octal to hexadecimal, or more precisely, they decided to add the ability to address 8-bit bytes as well as 6-bit bytes. That meant that they had six 8-bit nibbles in a word. That meant that they needed a hardware divide by 3 circuit in their nibble addressing. Divide by 3 isn't covered much in EE school.

The story is that they were interviewing a digital designer. They mentioned the problem with division by 3. He wrote down a solution for them. And then Burroughs didn't hire him because they didn't think he'd fit in with their corporate culture. But that's just the story.


[edit]The first Burroughs computer in the series was the B5000 and it's described here:

The requirement for a division by 3 was from the 6-bit addressing mentioned here:

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