Adding Hexadecimal NumbersDate: 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/ |
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