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:
B59e16 = B 5 9 e = B(=1011) 5(=0101) 9(=1001) e(=1110) = 10110101100111102
answer: B59e16 = 10110101100111102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙163+5∙162+9∙161+14∙160 = 11∙4096+5∙256+9∙16+14∙1 = 45056+1280+144+14 = 4649410
got It: B59e16 =4649410
Translate the number 4649410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
46494 | 2 | | | | | | | | | | | | | | | |
-46494 | 23247 | 2 | | | | | | | | | | | | | | |
0 | -23246 | 11623 | 2 | | | | | | | | | | | | | |
| 1 | -11622 | 5811 | 2 | | | | | | | | | | | | |
| | 1 | -5810 | 2905 | 2 | | | | | | | | | | | |
| | | 1 | -2904 | 1452 | 2 | | | | | | | | | | |
| | | | 1 | -1452 | 726 | 2 | | | | | | | | | |
| | | | | 0 | -726 | 363 | 2 | | | | | | | | |
| | | | | | 0 | -362 | 181 | 2 | | | | | | | |
| | | | | | | 1 | -180 | 90 | 2 | | | | | | |
| | | | | | | | 1 | -90 | 45 | 2 | | | | | |
| | | | | | | | | 0 | -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:
4649410 = 10110101100111102
answer: B59e16 = 10110101100111102