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:
7E0C16 = 7 E 0 C = 7(=0111) E(=1110) 0(=0000) C(=1100) = 1111110000011002
the Final answer: 7E0C16 = 1111110000011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
7∙163+14∙162+0∙161+12∙160 = 7∙4096+14∙256+0∙16+12∙1 = 28672+3584+0+12 = 3226810
got It: 7E0C16 =3226810
Translate the number 3226810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
32268 | 2 | | | | | | | | | | | | | | |
-32268 | 16134 | 2 | | | | | | | | | | | | | |
0 | -16134 | 8067 | 2 | | | | | | | | | | | | |
| 0 | -8066 | 4033 | 2 | | | | | | | | | | | |
| | 1 | -4032 | 2016 | 2 | | | | | | | | | | |
| | | 1 | -2016 | 1008 | 2 | | | | | | | | | |
| | | | 0 | -1008 | 504 | 2 | | | | | | | | |
| | | | | 0 | -504 | 252 | 2 | | | | | | | |
| | | | | | 0 | -252 | 126 | 2 | | | | | | |
| | | | | | | 0 | -126 | 63 | 2 | | | | | |
| | | | | | | | 0 | -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:
3226810 = 1111110000011002
the Final answer: 7E0C16 = 1111110000011002