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:
CABE16 = C A B E = C(=1100) A(=1010) B(=1011) E(=1110) = 11001010101111102
answer: CABE16 = 11001010101111102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙163+10∙162+11∙161+14∙160 = 12∙4096+10∙256+11∙16+14∙1 = 49152+2560+176+14 = 5190210
got It: CABE16 =5190210
Translate the number 5190210 в binary like this:
the Integer part of the number is divided by the base of the new number system:
51902 | 2 | | | | | | | | | | | | | | | |
-51902 | 25951 | 2 | | | | | | | | | | | | | | |
0 | -25950 | 12975 | 2 | | | | | | | | | | | | | |
| 1 | -12974 | 6487 | 2 | | | | | | | | | | | | |
| | 1 | -6486 | 3243 | 2 | | | | | | | | | | | |
| | | 1 | -3242 | 1621 | 2 | | | | | | | | | | |
| | | | 1 | -1620 | 810 | 2 | | | | | | | | | |
| | | | | 1 | -810 | 405 | 2 | | | | | | | | |
| | | | | | 0 | -404 | 202 | 2 | | | | | | | |
| | | | | | | 1 | -202 | 101 | 2 | | | | | | |
| | | | | | | | 0 | -100 | 50 | 2 | | | | | |
| | | | | | | | | 1 | -50 | 25 | 2 | | | | |
| | | | | | | | | | 0 | -24 | 12 | 2 | | | |
| | | | | | | | | | | 1 | -12 | 6 | 2 | | |
| | | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
5190210 = 11001010101111102
answer: CABE16 = 11001010101111102