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:
75CE16 = 7 5 C E = 7(=0111) 5(=0101) C(=1100) E(=1110) = 1110101110011102
the Final answer: 75CE16 = 1110101110011102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
7∙163+5∙162+12∙161+14∙160 = 7∙4096+5∙256+12∙16+14∙1 = 28672+1280+192+14 = 3015810
got It: 75CE16 =3015810
Translate the number 3015810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
30158 | 2 | | | | | | | | | | | | | | |
-30158 | 15079 | 2 | | | | | | | | | | | | | |
0 | -15078 | 7539 | 2 | | | | | | | | | | | | |
| 1 | -7538 | 3769 | 2 | | | | | | | | | | | |
| | 1 | -3768 | 1884 | 2 | | | | | | | | | | |
| | | 1 | -1884 | 942 | 2 | | | | | | | | | |
| | | | 0 | -942 | 471 | 2 | | | | | | | | |
| | | | | 0 | -470 | 235 | 2 | | | | | | | |
| | | | | | 1 | -234 | 117 | 2 | | | | | | |
| | | | | | | 1 | -116 | 58 | 2 | | | | | |
| | | | | | | | 1 | -58 | 29 | 2 | | | | |
| | | | | | | | | 0 | -28 | 14 | 2 | | | |
| | | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
3015810 = 1110101110011102
the Final answer: 75CE16 = 1110101110011102