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:
21BF16 = 2 1 B F = 2(=0010) 1(=0001) B(=1011) F(=1111) = 100001101111112
answer: 21BF16 = 100001101111112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+1∙162+11∙161+15∙160 = 2∙4096+1∙256+11∙16+15∙1 = 8192+256+176+15 = 863910
got It: 21BF16 =863910
Translate the number 863910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
8639 | 2 | | | | | | | | | | | | | |
-8638 | 4319 | 2 | | | | | | | | | | | | |
1 | -4318 | 2159 | 2 | | | | | | | | | | | |
| 1 | -2158 | 1079 | 2 | | | | | | | | | | |
| | 1 | -1078 | 539 | 2 | | | | | | | | | |
| | | 1 | -538 | 269 | 2 | | | | | | | | |
| | | | 1 | -268 | 134 | 2 | | | | | | | |
| | | | | 1 | -134 | 67 | 2 | | | | | | |
| | | | | | 0 | -66 | 33 | 2 | | | | | |
| | | | | | | 1 | -32 | 16 | 2 | | | | |
| | | | | | | | 1 | -16 | 8 | 2 | | | |
| | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
863910 = 100001101111112
answer: 21BF16 = 100001101111112