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:
5BDE16 = 5 B D E = 5(=0101) B(=1011) D(=1101) E(=1110) = 1011011110111102
answer: 5BDE16 = 1011011110111102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙163+11∙162+13∙161+14∙160 = 5∙4096+11∙256+13∙16+14∙1 = 20480+2816+208+14 = 2351810
got It: 5BDE16 =2351810
Translate the number 2351810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
23518 | 2 | | | | | | | | | | | | | | |
-23518 | 11759 | 2 | | | | | | | | | | | | | |
0 | -11758 | 5879 | 2 | | | | | | | | | | | | |
| 1 | -5878 | 2939 | 2 | | | | | | | | | | | |
| | 1 | -2938 | 1469 | 2 | | | | | | | | | | |
| | | 1 | -1468 | 734 | 2 | | | | | | | | | |
| | | | 1 | -734 | 367 | 2 | | | | | | | | |
| | | | | 0 | -366 | 183 | 2 | | | | | | | |
| | | | | | 1 | -182 | 91 | 2 | | | | | | |
| | | | | | | 1 | -90 | 45 | 2 | | | | | |
| | | | | | | | 1 | -44 | 22 | 2 | | | | |
| | | | | | | | | 1 | -22 | 11 | 2 | | | |
| | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
2351810 = 1011011110111102
answer: 5BDE16 = 1011011110111102