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can someone help me ?

  • 03-01-2012 03:39PM
    #1
    Registered Users, Registered Users 2 Posts: 1,651 ✭✭✭


    Hi im doing some study for a maths exam next week im stuck on one section and dont know how to work it out could some one help ?

    Its Converting different number base's in to other's i can do one side of the decimal point but not the fraction side of it.

    Eg
    2ED.4C to base 10 i can get the 2ED part but dont know hoe to work out the .4C


Comments

  • Registered Users, Registered Users 2 Posts: 1,651 ✭✭✭thenightrider


    I think i worked it out but not 100 % sure 749.296875 :confused:


  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    I got the same answer as you. The first digit after the "decimal" point is sixteenths, the next is 256ths, and so on.

    2ED.4C is therefore: 2*4096 + 14*256 + 13*16 + 4*(1/16) + 12 * (1/256)


  • Registered Users, Registered Users 2 Posts: 13,183 ✭✭✭✭bnt


    Aoother way to do this is:
    - make note of the number of fraction digits (f) behind the decimal point (f = 2)
    - convert without the decimal point (2ED4Ch => 191820)
    - divide the result by base^f: 191820 / (16^2) = 191820/256 = 749.296875

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  • Registered Users, Registered Users 2 Posts: 1,501 ✭✭✭Delphi91


    Another approach:

    Count the number of characters BEFORE the decimal point (n=3).

    Multiply the first character by 16^(n-1), the second by 16^(n-2) etc until you get to the end. Evaluate each term and add them up i.e.

    [latex]\displaystyle {2 \times 16^2} + {E \times 16^1} + {D \times 16^0} + {4 \times 16^{-1}} + {C \times 16^{-2}}\\ [/latex]

    Converting the letters to their respective numbers gives:

    [latex]\displaystyle = 2 \times 256 + 14 \times 16 + 13 \times 1 + 4 \times 0.0625 + 12 \times 0.0039\\ [/latex]

    [latex]\displaystyle $$= 512 $ + $ 240 $ + $ 13 $ + $ 0.25 $ + $ 0.0468$$\\ [/latex]

    [latex]\displaystyle = 765.2968_{10}[/latex]


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