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:
75EC16 = 7 5 E C = 7(=0111) 5(=0101) E(=1110) C(=1100) = 1110101111011002
the Final answer: 75EC16 = 1110101111011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
7∙163+5∙162+14∙161+12∙160 = 7∙4096+5∙256+14∙16+12∙1 = 28672+1280+224+12 = 3018810
got It: 75EC16 =3018810
Translate the number 3018810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
30188 | 2 | | | | | | | | | | | | | | |
-30188 | 15094 | 2 | | | | | | | | | | | | | |
0 | -15094 | 7547 | 2 | | | | | | | | | | | | |
| 0 | -7546 | 3773 | 2 | | | | | | | | | | | |
| | 1 | -3772 | 1886 | 2 | | | | | | | | | | |
| | | 1 | -1886 | 943 | 2 | | | | | | | | | |
| | | | 0 | -942 | 471 | 2 | | | | | | | | |
| | | | | 1 | -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:
3018810 = 1110101111011002
the Final answer: 75EC16 = 1110101111011002