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:
2bad16 = 2 b a d = 2(=0010) b(=1011) a(=1010) d(=1101) = 101011101011012
answer: 2bad16 = 101011101011012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+11∙162+10∙161+13∙160 = 2∙4096+11∙256+10∙16+13∙1 = 8192+2816+160+13 = 1118110
got It: 2bad16 =1118110
Translate the number 1118110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
11181 | 2 | | | | | | | | | | | | | |
-11180 | 5590 | 2 | | | | | | | | | | | | |
1 | -5590 | 2795 | 2 | | | | | | | | | | | |
| 0 | -2794 | 1397 | 2 | | | | | | | | | | |
| | 1 | -1396 | 698 | 2 | | | | | | | | | |
| | | 1 | -698 | 349 | 2 | | | | | | | | |
| | | | 0 | -348 | 174 | 2 | | | | | | | |
| | | | | 1 | -174 | 87 | 2 | | | | | | |
| | | | | | 0 | -86 | 43 | 2 | | | | | |
| | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
1118110 = 101011101011012
answer: 2bad16 = 101011101011012