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:
ff7f16 = f f 7 f = f(=1111) f(=1111) 7(=0111) f(=1111) = 11111111011111112
answer: ff7f16 = 11111111011111112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
15∙163+15∙162+7∙161+15∙160 = 15∙4096+15∙256+7∙16+15∙1 = 61440+3840+112+15 = 6540710
got It: ff7f16 =6540710
Translate the number 6540710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
65407 | 2 | | | | | | | | | | | | | | | |
-65406 | 32703 | 2 | | | | | | | | | | | | | | |
1 | -32702 | 16351 | 2 | | | | | | | | | | | | | |
| 1 | -16350 | 8175 | 2 | | | | | | | | | | | | |
| | 1 | -8174 | 4087 | 2 | | | | | | | | | | | |
| | | 1 | -4086 | 2043 | 2 | | | | | | | | | | |
| | | | 1 | -2042 | 1021 | 2 | | | | | | | | | |
| | | | | 1 | -1020 | 510 | 2 | | | | | | | | |
| | | | | | 1 | -510 | 255 | 2 | | | | | | | |
| | | | | | | 0 | -254 | 127 | 2 | | | | | | |
| | | | | | | | 1 | -126 | 63 | 2 | | | | | |
| | | | | | | | | 1 | -62 | 31 | 2 | | | | |
| | | | | | | | | | 1 | -30 | 15 | 2 | | | |
| | | | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
6540710 = 11111111011111112
answer: ff7f16 = 11111111011111112