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:
3BAD16 = 3 B A D = 3(=0011) B(=1011) A(=1010) D(=1101) = 111011101011012
answer: 3BAD16 = 111011101011012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+11∙162+10∙161+13∙160 = 3∙4096+11∙256+10∙16+13∙1 = 12288+2816+160+13 = 1527710
got It: 3BAD16 =1527710
Translate the number 1527710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
15277 | 2 | | | | | | | | | | | | | |
-15276 | 7638 | 2 | | | | | | | | | | | | |
1 | -7638 | 3819 | 2 | | | | | | | | | | | |
| 0 | -3818 | 1909 | 2 | | | | | | | | | | |
| | 1 | -1908 | 954 | 2 | | | | | | | | | |
| | | 1 | -954 | 477 | 2 | | | | | | | | |
| | | | 0 | -476 | 238 | 2 | | | | | | | |
| | | | | 1 | -238 | 119 | 2 | | | | | | |
| | | | | | 0 | -118 | 59 | 2 | | | | | |
| | | | | | | 1 | -58 | 29 | 2 | | | | |
| | | | | | | | 1 | -28 | 14 | 2 | | | |
| | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1527710 = 111011101011012
answer: 3BAD16 = 111011101011012