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Adding Hexadecimal Numbers

Date: 01/31/2005 at 22:16:12
From: Hitesh
Subject: Adding Hexadecimals that are complex

I'm trying to add two hexadecimal numbers, ACF1 + AFFF.  I've read 
your guides on this website and I still can't get an idea about how I
should go about working out this problem.  I'm not looking for the
answer, I'm just looking for a push in the right direction if 
possible.



Date: 02/01/2005 at 03:15:11
From: Doctor Achilles
Subject: Re: Adding Hexadecimals that are complex

Hi Hitesh,

Thanks for writing to Dr. Math.

I have to admit I haven't read all of our archives on hex, so I
appologize if this is repetitious, but I want to start from scratch.

One-digit hex addition (I will omit examples of the commutative
property such as 1+2 = 2+1):

 1+1 = 2
 1+2 = 3
 1+3 = 2+2 = 4
 1+4 = 2+3 = 5
 1+5 = 2+4 = 3+3 = 6
 1+6 = 2+5 = 3+4 = 7
 1+7 = 2+6 = 3+5 = 4+4 = 8
 1+8 = 2+7 = 3+6 = 4+5 = 9
 1+9 = 2+8 = 3+7 = 4+6 = 5+5 = A
 1+A = 2+9 = 3+8 = 4+7 = 5+6 = B
 1+B = 2+A = 3+9 = 4+8 = 5+7 = 6+6 = C
 1+C = 2+B = 3+A = 4+9 = 5+8 = 6+7 = D
 1+D = 2+C = 3+B = 4+A = 5+9 = 6+8 = 7+7 = E
 1+E = 2+D = 3+C = 4+B = 5+A = 6+9 = 7+8 = F
 1+F = 2+E = 3+D = 4+C = 5+B = 6+A = 7+9 = 8+8 = 10
 2+F = 3+E = 4+D = 5+C = 6+B = 7+A = 8+9 = 11
 3+F = 4+E = 5+D = 6+C = 7+B = 8+A = 9+9 = 12
 4+F = 5+E = 6+D = 7+C = 8+B = 9+A = 13
 5+F = 6+E = 7+D = 8+C = 9+B = A+A = 14
 6+F = 7+E = 8+D = 9+C = A+B = 15
 7+F = 8+E = 9+D = A+C = B+B = 16
 8+F = 9+E = A+D = B+C = 17
 9+F = A+E = B+D = C+C = 18
 A+F = B+E = C+D = 19
 B+F = C+E = D+D = 1A
 C+F = D+E = 1B
 D+F = E+E = 1C
 E+F = 1D
 F+F = 1E

Ok, now to do addition, you do it just like normal addition.  Line the
numbers up, start by adding the one's digit and carry the 1 if there
is one.

E.g.

    182AB
  + 5FCAA
 ---------

First, add the one's digit: B+A = 15, so I carry the 1:

       1
    182AB
  + 5FCAA
 ---------
        5

Now, add the ten's digit: A+A = 14, plus the 1 I carried = 15. 
Remember to carry the 1 again:

      11
    182AB
  + 5FCAA
 ---------
       55

Now, add the hundred's digit: 2+C = E, plus the 1 I carried = F:

      11
    182AB
  + 5FCAA
 ---------
      F55

Now, add the thousands digit: 8+F = 17, carry the 1:

    1 11
    182AB
  + 5FCAA
 ---------
     7F55

Now, add the last digit: 1+5 = 6, plus the 1 I carried = 7:

    1 11
    182AB
  + 5FCAA
 ---------
    77F55

So:

  182AB + 5FCAA = 77F55

Hope this helps.  If you have other questions or you'd like to talk
about this some more, please write back.

- Doctor Achilles, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Number Theory

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