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:
AF516 = A F 5 = A(=1010) F(=1111) 5(=0101) = 1010111101012
the Final answer: AF516 = 1010111101012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙162+15∙161+5∙160 = 10∙256+15∙16+5∙1 = 2560+240+5 = 280510
got It: AF516 =280510
Translate the number 280510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2805 | 2 | | | | | | | | | | | |
-2804 | 1402 | 2 | | | | | | | | | | |
1 | -1402 | 701 | 2 | | | | | | | | | |
| 0 | -700 | 350 | 2 | | | | | | | | |
| | 1 | -350 | 175 | 2 | | | | | | | |
| | | 0 | -174 | 87 | 2 | | | | | | |
| | | | 1 | -86 | 43 | 2 | | | | | |
| | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | 0 | | |
|
the result of the conversion was:
280510 = 1010111101012
the Final answer: AF516 = 1010111101012