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:
AFB16 = A F B = A(=1010) F(=1111) B(=1011) = 1010111110112
the Final answer: AFB16 = 1010111110112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙162+15∙161+11∙160 = 10∙256+15∙16+11∙1 = 2560+240+11 = 281110
got It: AFB16 =281110
Translate the number 281110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2811 | 2 | | | | | | | | | | | |
-2810 | 1405 | 2 | | | | | | | | | | |
1 | -1404 | 702 | 2 | | | | | | | | | |
| 1 | -702 | 351 | 2 | | | | | | | | |
| | 0 | -350 | 175 | 2 | | | | | | | |
| | | 1 | -174 | 87 | 2 | | | | | | |
| | | | 1 | -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:
281110 = 1010111110112
the Final answer: AFB16 = 1010111110112