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:
A94F16 = A 9 4 F = A(=1010) 9(=1001) 4(=0100) F(=1111) = 10101001010011112
answer: A94F16 = 10101001010011112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+9∙162+4∙161+15∙160 = 10∙4096+9∙256+4∙16+15∙1 = 40960+2304+64+15 = 4334310
got It: A94F16 =4334310
Translate the number 4334310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
43343 | 2 | | | | | | | | | | | | | | | |
-43342 | 21671 | 2 | | | | | | | | | | | | | | |
1 | -21670 | 10835 | 2 | | | | | | | | | | | | | |
| 1 | -10834 | 5417 | 2 | | | | | | | | | | | | |
| | 1 | -5416 | 2708 | 2 | | | | | | | | | | | |
| | | 1 | -2708 | 1354 | 2 | | | | | | | | | | |
| | | | 0 | -1354 | 677 | 2 | | | | | | | | | |
| | | | | 0 | -676 | 338 | 2 | | | | | | | | |
| | | | | | 1 | -338 | 169 | 2 | | | | | | | |
| | | | | | | 0 | -168 | 84 | 2 | | | | | | |
| | | | | | | | 1 | -84 | 42 | 2 | | | | | |
| | | | | | | | | 0 | -42 | 21 | 2 | | | | |
| | | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4334310 = 10101001010011112
answer: A94F16 = 10101001010011112