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:
0b111116 = 0 b 1 1 1 1 = 0(=0000) b(=1011) 1(=0001) 1(=0001) 1(=0001) 1(=0001) = 101100010001000100012
answer: 0b111116 = 101100010001000100012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙165+11∙164+1∙163+1∙162+1∙161+1∙160 = 0∙1048576+11∙65536+1∙4096+1∙256+1∙16+1∙1 = 0+720896+4096+256+16+1 = 72526510
got It: 0b111116 =72526510
Translate the number 72526510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
725265 | 2 | | | | | | | | | | | | | | | | | | | |
-725264 | 362632 | 2 | | | | | | | | | | | | | | | | | | |
1 | -362632 | 181316 | 2 | | | | | | | | | | | | | | | | | |
| 0 | -181316 | 90658 | 2 | | | | | | | | | | | | | | | | |
| | 0 | -90658 | 45329 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -45328 | 22664 | 2 | | | | | | | | | | | | | | |
| | | | 1 | -22664 | 11332 | 2 | | | | | | | | | | | | | |
| | | | | 0 | -11332 | 5666 | 2 | | | | | | | | | | | | |
| | | | | | 0 | -5666 | 2833 | 2 | | | | | | | | | | | |
| | | | | | | 0 | -2832 | 1416 | 2 | | | | | | | | | | |
| | | | | | | | 1 | -1416 | 708 | 2 | | | | | | | | | |
| | | | | | | | | 0 | -708 | 354 | 2 | | | | | | | | |
| | | | | | | | | | 0 | -354 | 177 | 2 | | | | | | | |
| | | | | | | | | | | 0 | -176 | 88 | 2 | | | | | | |
| | | | | | | | | | | | 1 | -88 | 44 | 2 | | | | | |
| | | | | | | | | | | | | 0 | -44 | 22 | 2 | | | | |
| | | | | | | | | | | | | | 0 | -22 | 11 | 2 | | | |
| | | | | | | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
72526510 = 101100010001000100012
answer: 0b111116 = 101100010001000100012