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:
1AB316 = 1 A B 3 = 1(=0001) A(=1010) B(=1011) 3(=0011) = 11010101100112
answer: 1AB316 = 11010101100112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙163+10∙162+11∙161+3∙160 = 1∙4096+10∙256+11∙16+3∙1 = 4096+2560+176+3 = 683510
got It: 1AB316 =683510
Translate the number 683510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
6835 | 2 | | | | | | | | | | | | |
-6834 | 3417 | 2 | | | | | | | | | | | |
1 | -3416 | 1708 | 2 | | | | | | | | | | |
| 1 | -1708 | 854 | 2 | | | | | | | | | |
| | 0 | -854 | 427 | 2 | | | | | | | | |
| | | 0 | -426 | 213 | 2 | | | | | | | |
| | | | 1 | -212 | 106 | 2 | | | | | | |
| | | | | 1 | -106 | 53 | 2 | | | | | |
| | | | | | 0 | -52 | 26 | 2 | | | | |
| | | | | | | 1 | -26 | 13 | 2 | | | |
| | | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | 1 | | |
|
the result of the conversion was:
683510 = 11010101100112
answer: 1AB316 = 11010101100112