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A question on changing bases of numbers

  1. Feb 3, 2008 #1
    i want to transform a number in base 6 into a number in base 3


    i was looking for a way to transform straight forward without any middle man
    ("from 6 to 10 and from 10 to 3" thats not the way i am looking for)
     
  2. jcsd
  3. Feb 3, 2008 #2

    EnumaElish

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    Are you looking for a calculator (see http://www.1728.com/numbbase.htm) or a formula?

    The algorithm for conversion into decimal is:

    http://mathforum.org/library/drmath/view/57074.html

    In base 5 this is: 97 = 3x25 + 4x5 + 2x1 = 3425.

    If you wanted to convert 101213 to base 5 directly, you'd need to figure out the "place values" in base 5.

    {1, 3, 9, 27, 81} = {15, 35, 145, 1025, 3115}

    Then: 1 *3115 + 0 * 1025 + 1 * 145 + 2* 35 + 1* 15 = (3115 + 145) + (115 + 15) = 3305 + 125 = 3425
     
    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    i am looking for a formula
     
  5. Feb 3, 2008 #4

    EnumaElish

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    See edits to my post.
     
  6. Feb 3, 2008 #5
    i know how to convert to ten base from each number
    i want to covert from base 6 to base 3 straight forward
     
  7. May 7, 2009 #6
    Easy! Just learn how to add, subtract, multiply, and divide in base 6. Doing so with "3" is very easy, because it is just half of 6 (kind of like working with "5" in base ten). Then convert the exact same way you otherwise would - only in base six. Ex:

    12345 base six should be 1010,20111 base three (check me)
     
  8. May 7, 2009 #7
    I'm sorry! I actually wound up calculating the REVERSE sequence: "54321 base six"!

    For reference, please let me show you how I arrived at this number - going straight from base six to base three:

    15304 base six
    ______
    3)54321 base six

    Remainder: 1

    3501 base six
    ______
    3)15304 base six ... (I will cease to make this note, and let any following number be base six unless otherwise noted)

    Remainder: 1

    1140
    _____
    3)3501

    Remainder: 1

    232
    _____
    3)1140

    Remainder: 0

    50
    ____
    3)232

    Remainder: 2

    14
    ___
    3)50

    Remainder: 0

    3
    ___
    3)14

    Remainder: 1

    1
    __
    3)3

    Remainder: 0

    0
    __
    3)1

    Remainder: 1

    Now, put the "remainders" in order, starting with the last and ending with the first:

    1010,20111 base three (trinary)

    Now, let's try "12345 base six":

    2513
    ______
    3)12345 base six (as will be the case UON (unless otherwise noted))
    -10
    23
    -23
    04
    -3
    15
    -13
    Remainder: 2

    543
    _____
    3)2513
    -23
    21
    -20
    13
    -13
    Remainder: 0

    153
    ____
    3)543
    -3
    24
    -23
    13
    -13
    Remainder: 0

    35
    ____
    3)153
    -13
    23
    -23
    Remainder: 0

    11
    ___
    3)35
    -33
    Remainder: 2

    2
    ___
    3)11
    -10
    Remainder: 1

    0
    __
    3)2
    Remainder: 2

    Now, again, just arrange the respective remainders in reverse order for the trinary equivalent: 2120002 base three (trinary)

    (By the way, I have come to prefer grouping my numerical expressions into groups of five bits, instead of three; as you may have noticed. My reason for this preference would actually be quite a long story).

    doGraeF
     
  9. May 29, 2009 #8
    An even better answer!

    I found an even more convenient method of changing bases. Let's consider base six: "12345" again.

    First, we take the left-most digit: "1". Then, if it were "3" or higher, we would convert it to base three, but "1" is the same in base six as in base three.

    Second, multiply it by two: 1*2 = 2

    Third, multiple by base three "10": 2*10 = 20 base three


    Now, we take the second left-most digit: "2", and leave it alone, because it is the same in base six or base three. Then, add it to our running total: 20+2 = 22.

    Multiply 22 by 2: 22 * 2 = 110 + 11 = 121.

    And multiply 121 by 10: 121 * 10 = 1210.


    Now, we take the middle digit: "3", and convert it to base three: "10". Add it to 1210: 1210 + 10 = 1220.

    Multiply by 2: 1220 * 2 = 2000 + 1100 + 110 = 10210.

    Multiply by 10: 10210 * 10 = 1,02100.


    Take second right-most digit: "4", and convert to base three: "11". Add it to 102100: 1,02100 + 11 = 1,02111.

    1,02111 * 2 = 2,11222.

    2,11222 * 10 = 21,12220.


    Take the right-most digit: "5", and convert to base three: "12". Add it to 21,12220: 21,12220 + 12: 21,20002.

    So, in summary: work from left to right, take the the digit, convert to base three, multiply by 2, then multiply by base three "10", then convert the next digit to base three and add it to the running total, multiply the running total by 2, multply the running total by base three "10", and etc. until you have finally added the rightmost digit, converted to base three, to the running total. At that point, the running total will be your base six number converted to base three.

    Daniel
     
  10. May 30, 2009 #9
    As if I have not already said enough about this (and do bear with me if this seems a little excessive), let's try our other example: 54321 base six. (All numbers will be in base three unless otherwise noted).

    First, let's take the left-most digit: "5", and convert it to base three: "12". Now, let's multiply it by two: 12 * 2 = 20 + 11 = 101. Next, multiply it by 10: 101 * 10 = 1010.

    Now, we convert the second left-most digit, "4", to base three: 11. Now, add it to our running total: 1010 + 11 = 1021. Multiply running total by 2: 1021 * 2 = 2000 + 110 + 2 = 2112. Multiply it by 10: 2112 + 10 = 21120.

    Take the middle digit, "3", and convert to base three: 10. Add it to running total: 21120 + 10 = 21200. Multiply total by 2: 21200 * 2 = 1,10000 + 2000 + 1100 = 1,20100. Multiply it by 10: 1,20100 * 10 + 12,01000.

    Add second right-most digit, "2" to total: 12,01000 + 2 = 12,01002. Multiply by 2: 12,01002 * 2 = 20,00000 + 11,00000 + 2000 + 11 = 101,02011. Multiply by 10: 101,02011 * 10 = 1010,02110.

    Add right-most digit, "1", to total: 1010,02110 + 1 = 1010,02111.
     
    Last edited: May 30, 2009
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