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:
FA9F16 = F A 9 F = F(=1111) A(=1010) 9(=1001) F(=1111) = 11111010100111112
answer: FA9F16 = 11111010100111112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
15∙163+10∙162+9∙161+15∙160 = 15∙4096+10∙256+9∙16+15∙1 = 61440+2560+144+15 = 6415910
got It: FA9F16 =6415910
Translate the number 6415910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
64159 | 2 | | | | | | | | | | | | | | | |
-64158 | 32079 | 2 | | | | | | | | | | | | | | |
1 | -32078 | 16039 | 2 | | | | | | | | | | | | | |
| 1 | -16038 | 8019 | 2 | | | | | | | | | | | | |
| | 1 | -8018 | 4009 | 2 | | | | | | | | | | | |
| | | 1 | -4008 | 2004 | 2 | | | | | | | | | | |
| | | | 1 | -2004 | 1002 | 2 | | | | | | | | | |
| | | | | 0 | -1002 | 501 | 2 | | | | | | | | |
| | | | | | 0 | -500 | 250 | 2 | | | | | | | |
| | | | | | | 1 | -250 | 125 | 2 | | | | | | |
| | | | | | | | 0 | -124 | 62 | 2 | | | | | |
| | | | | | | | | 1 | -62 | 31 | 2 | | | | |
| | | | | | | | | | 0 | -30 | 15 | 2 | | | |
| | | | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
6415910 = 11111010100111112
answer: FA9F16 = 11111010100111112