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:
1A9e16 = 1 A 9 e = 1(=0001) A(=1010) 9(=1001) e(=1110) = 11010100111102
answer: 1A9e16 = 11010100111102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙163+10∙162+9∙161+14∙160 = 1∙4096+10∙256+9∙16+14∙1 = 4096+2560+144+14 = 681410
got It: 1A9e16 =681410
Translate the number 681410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
6814 | 2 | | | | | | | | | | | | |
-6814 | 3407 | 2 | | | | | | | | | | | |
0 | -3406 | 1703 | 2 | | | | | | | | | | |
| 1 | -1702 | 851 | 2 | | | | | | | | | |
| | 1 | -850 | 425 | 2 | | | | | | | | |
| | | 1 | -424 | 212 | 2 | | | | | | | |
| | | | 1 | -212 | 106 | 2 | | | | | | |
| | | | | 0 | -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:
681410 = 11010100111102
answer: 1A9e16 = 11010100111102