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:
A3D416 = A 3 D 4 = A(=1010) 3(=0011) D(=1101) 4(=0100) = 10100011110101002
answer: A3D416 = 10100011110101002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+3∙162+13∙161+4∙160 = 10∙4096+3∙256+13∙16+4∙1 = 40960+768+208+4 = 4194010
got It: A3D416 =4194010
Translate the number 4194010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
41940 | 2 | | | | | | | | | | | | | | | |
-41940 | 20970 | 2 | | | | | | | | | | | | | | |
0 | -20970 | 10485 | 2 | | | | | | | | | | | | | |
| 0 | -10484 | 5242 | 2 | | | | | | | | | | | | |
| | 1 | -5242 | 2621 | 2 | | | | | | | | | | | |
| | | 0 | -2620 | 1310 | 2 | | | | | | | | | | |
| | | | 1 | -1310 | 655 | 2 | | | | | | | | | |
| | | | | 0 | -654 | 327 | 2 | | | | | | | | |
| | | | | | 1 | -326 | 163 | 2 | | | | | | | |
| | | | | | | 1 | -162 | 81 | 2 | | | | | | |
| | | | | | | | 1 | -80 | 40 | 2 | | | | | |
| | | | | | | | | 1 | -40 | 20 | 2 | | | | |
| | | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4194010 = 10100011110101002
answer: A3D416 = 10100011110101002