This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from hexadecimal to binary like this:
DB4F16 = D B 4 F = D(=1101) B(=1011) 4(=0100) F(=1111) = 11011011010011112
answer: DB4F16 = 11011011010011112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
13∙163+11∙162+4∙161+15∙160 = 13∙4096+11∙256+4∙16+15∙1 = 53248+2816+64+15 = 5614310
got It: DB4F16 =5614310
Translate the number 5614310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
56143 | 2 | | | | | | | | | | | | | | | |
-56142 | 28071 | 2 | | | | | | | | | | | | | | |
1 | -28070 | 14035 | 2 | | | | | | | | | | | | | |
| 1 | -14034 | 7017 | 2 | | | | | | | | | | | | |
| | 1 | -7016 | 3508 | 2 | | | | | | | | | | | |
| | | 1 | -3508 | 1754 | 2 | | | | | | | | | | |
| | | | 0 | -1754 | 877 | 2 | | | | | | | | | |
| | | | | 0 | -876 | 438 | 2 | | | | | | | | |
| | | | | | 1 | -438 | 219 | 2 | | | | | | | |
| | | | | | | 0 | -218 | 109 | 2 | | | | | | |
| | | | | | | | 1 | -108 | 54 | 2 | | | | | |
| | | | | | | | | 1 | -54 | 27 | 2 | | | | |
| | | | | | | | | | 0 | -26 | 13 | 2 | | | |
| | | | | | | | | | | 1 | -12 | 6 | 2 | | |
| | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
5614310 = 11011011010011112
answer: DB4F16 = 11011011010011112