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:
A44D16 = A 4 4 D = A(=1010) 4(=0100) 4(=0100) D(=1101) = 10100100010011012
answer: A44D16 = 10100100010011012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+4∙162+4∙161+13∙160 = 10∙4096+4∙256+4∙16+13∙1 = 40960+1024+64+13 = 4206110
got It: A44D16 =4206110
Translate the number 4206110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
42061 | 2 | | | | | | | | | | | | | | | |
-42060 | 21030 | 2 | | | | | | | | | | | | | | |
1 | -21030 | 10515 | 2 | | | | | | | | | | | | | |
| 0 | -10514 | 5257 | 2 | | | | | | | | | | | | |
| | 1 | -5256 | 2628 | 2 | | | | | | | | | | | |
| | | 1 | -2628 | 1314 | 2 | | | | | | | | | | |
| | | | 0 | -1314 | 657 | 2 | | | | | | | | | |
| | | | | 0 | -656 | 328 | 2 | | | | | | | | |
| | | | | | 1 | -328 | 164 | 2 | | | | | | | |
| | | | | | | 0 | -164 | 82 | 2 | | | | | | |
| | | | | | | | 0 | -82 | 41 | 2 | | | | | |
| | | | | | | | | 0 | -40 | 20 | 2 | | | | |
| | | | | | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4206110 = 10100100010011012
answer: A44D16 = 10100100010011012