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:
A0F116 = A 0 F 1 = A(=1010) 0(=0000) F(=1111) 1(=0001) = 10100000111100012
answer: A0F116 = 10100000111100012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+0∙162+15∙161+1∙160 = 10∙4096+0∙256+15∙16+1∙1 = 40960+0+240+1 = 4120110
got It: A0F116 =4120110
Translate the number 4120110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
41201 | 2 | | | | | | | | | | | | | | | |
-41200 | 20600 | 2 | | | | | | | | | | | | | | |
1 | -20600 | 10300 | 2 | | | | | | | | | | | | | |
| 0 | -10300 | 5150 | 2 | | | | | | | | | | | | |
| | 0 | -5150 | 2575 | 2 | | | | | | | | | | | |
| | | 0 | -2574 | 1287 | 2 | | | | | | | | | | |
| | | | 1 | -1286 | 643 | 2 | | | | | | | | | |
| | | | | 1 | -642 | 321 | 2 | | | | | | | | |
| | | | | | 1 | -320 | 160 | 2 | | | | | | | |
| | | | | | | 1 | -160 | 80 | 2 | | | | | | |
| | | | | | | | 0 | -80 | 40 | 2 | | | | | |
| | | | | | | | | 0 | -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:
4120110 = 10100000111100012
answer: A0F116 = 10100000111100012