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Octal Multiplication confusion 
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#1
Oct905, 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. http://img206.imageshack.us/img206/6219/lastscan8oo.jpg 


#2
Oct905, 11:47 AM

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Wow, that's bad notation. 10 is not equal to 12. 10_{10} = 12[/sub]8[/sub].
8_{10} + 2_{10} = (1 x 8^{1}) + (2 x 8^{0}) = 12_{8} 24_{10} + 7_{10} = (3 x 8^{1}) + (7 x 8^{0}) = 37_{8} Do you get it now? For comparison, consider: 1000 + 200 + 30 + 4 = (1 x 10³) + (2 x 10²) + (3 x 10^{1}) + (4 x 10^{0}) = 1234_{10} 


#3
Oct905, 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. Thanks, Tom 


#4
Oct905, 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!



#5
Oct1005, 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: 00762 00045 _____ becomes 000,000,111,110,010 000,000,000,100,101 ___________________ 000,000,111,110,010 000,011,111,001,0 011,111,001,0 ___________________ 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: 43772 A computer would do the multiplication in binary simply by  shift and add  shift and add  etc. KM 


#6
Oct1105, 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 48bit word which consisted of 16 octal digits. They could address by words, and they also could address the eight 6bit 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 8bit bytes as well as 6bit bytes. That meant that they had six 8bit 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. Carl [edit]The first Burroughs computer in the series was the B5000 and it's described here: http://www.answers.com/topic/burroughsb5000 The requirement for a division by 3 was from the 6bit addressing mentioned here: http://en.wikipedia.org/wiki/B5000_Descriptors [/edit] 


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