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:
507F16 = 5 0 7 F = 5(=0101) 0(=0000) 7(=0111) F(=1111) = 1010000011111112
answer: 507F16 = 1010000011111112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙163+0∙162+7∙161+15∙160 = 5∙4096+0∙256+7∙16+15∙1 = 20480+0+112+15 = 2060710
got It: 507F16 =2060710
Translate the number 2060710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
20607 | 2 | | | | | | | | | | | | | | |
-20606 | 10303 | 2 | | | | | | | | | | | | | |
1 | -10302 | 5151 | 2 | | | | | | | | | | | | |
| 1 | -5150 | 2575 | 2 | | | | | | | | | | | |
| | 1 | -2574 | 1287 | 2 | | | | | | | | | | |
| | | 1 | -1286 | 643 | 2 | | | | | | | | | |
| | | | 1 | -642 | 321 | 2 | | | | | | | | |
| | | | | 1 | -320 | 160 | 2 | | | | | | | |
| | | | | | 1 | -160 | 80 | 2 | | | | | | |
| | | | | | | 0 | -80 | 40 | 2 | | | | | |
| | | | | | | | 0 | -40 | 20 | 2 | | | | |
| | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
2060710 = 1010000011111112
answer: 507F16 = 1010000011111112