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In BCD addition why add 6?

  1. Sep 29, 2012 #1
    In BCD addition when the number exceed 9 we add 6 as the correction value. Although I tried all the cases and saw tthat 6 was indeed the asnwer, how do I prove that? Suppose that instead of BCD I had a 6-bit number, what would be the correction value for that coding?
     
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  3. Oct 5, 2012 #2

    berkeman

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    Staff: Mentor

    http://en.wikipedia.org/wiki/Binary-coded_decimal

    When you get to 0x09, what is the binary representation of it? And to get to 0x10, what do you have to add to 0x09?
     
  4. Oct 6, 2012 #3
    What is x in 0x09. Anyways for 9 representation is 1001 and for 10 its 1010 in binary.
    In BCD representation for 9 is the same but for 10 its 0001 0000 instead of 1010.
    We need to add 6 for that - I get that but how to prove it?
     
  5. Oct 6, 2012 #4

    berkeman

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    Staff: Mentor

    Standard notation. 0x = hex, 0b = binary...
     
  6. Oct 6, 2012 #5
    The max amount of digits that a decimal increments before a carry is 9. The four bits used in the BCD nibble allows the max value of 16 to be counted. Therefore to properly carry from one nibble to the next 6 is added to any value above 9 to implement a carry to the next nibble. As for the the six bit number I would assume it would be the difference between the max size of the amount of bits you are using and 9.
    ((2^n)-1)-9= the amount needed to carry into the next group of bits
    where n is the number of bits used.
     
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