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:
ABC716 = A B C 7 = A(=1010) B(=1011) C(=1100) 7(=0111) = 10101011110001112
answer: ABC716 = 10101011110001112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+11∙162+12∙161+7∙160 = 10∙4096+11∙256+12∙16+7∙1 = 40960+2816+192+7 = 4397510
got It: ABC716 =4397510
Translate the number 4397510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
43975 | 2 | | | | | | | | | | | | | | | |
-43974 | 21987 | 2 | | | | | | | | | | | | | | |
1 | -21986 | 10993 | 2 | | | | | | | | | | | | | |
| 1 | -10992 | 5496 | 2 | | | | | | | | | | | | |
| | 1 | -5496 | 2748 | 2 | | | | | | | | | | | |
| | | 0 | -2748 | 1374 | 2 | | | | | | | | | | |
| | | | 0 | -1374 | 687 | 2 | | | | | | | | | |
| | | | | 0 | -686 | 343 | 2 | | | | | | | | |
| | | | | | 1 | -342 | 171 | 2 | | | | | | | |
| | | | | | | 1 | -170 | 85 | 2 | | | | | | |
| | | | | | | | 1 | -84 | 42 | 2 | | | | | |
| | | | | | | | | 1 | -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:
4397510 = 10101011110001112
answer: ABC716 = 10101011110001112