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:
Bad116 = B a d 1 = B(=1011) a(=1010) d(=1101) 1(=0001) = 10111010110100012
answer: Bad116 = 10111010110100012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙163+10∙162+13∙161+1∙160 = 11∙4096+10∙256+13∙16+1∙1 = 45056+2560+208+1 = 4782510
got It: Bad116 =4782510
Translate the number 4782510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
47825 | 2 | | | | | | | | | | | | | | | |
-47824 | 23912 | 2 | | | | | | | | | | | | | | |
1 | -23912 | 11956 | 2 | | | | | | | | | | | | | |
| 0 | -11956 | 5978 | 2 | | | | | | | | | | | | |
| | 0 | -5978 | 2989 | 2 | | | | | | | | | | | |
| | | 0 | -2988 | 1494 | 2 | | | | | | | | | | |
| | | | 1 | -1494 | 747 | 2 | | | | | | | | | |
| | | | | 0 | -746 | 373 | 2 | | | | | | | | |
| | | | | | 1 | -372 | 186 | 2 | | | | | | | |
| | | | | | | 1 | -186 | 93 | 2 | | | | | | |
| | | | | | | | 0 | -92 | 46 | 2 | | | | | |
| | | | | | | | | 1 | -46 | 23 | 2 | | | | |
| | | | | | | | | | 0 | -22 | 11 | 2 | | | |
| | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4782510 = 10111010110100012
answer: Bad116 = 10111010110100012