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:
9BC816 = 9 B C 8 = 9(=1001) B(=1011) C(=1100) 8(=1000) = 10011011110010002
the Final answer: 9BC816 = 10011011110010002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
9∙163+11∙162+12∙161+8∙160 = 9∙4096+11∙256+12∙16+8∙1 = 36864+2816+192+8 = 3988010
got It: 9BC816 =3988010
Translate the number 3988010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
39880 | 2 | | | | | | | | | | | | | | | |
-39880 | 19940 | 2 | | | | | | | | | | | | | | |
0 | -19940 | 9970 | 2 | | | | | | | | | | | | | |
| 0 | -9970 | 4985 | 2 | | | | | | | | | | | | |
| | 0 | -4984 | 2492 | 2 | | | | | | | | | | | |
| | | 1 | -2492 | 1246 | 2 | | | | | | | | | | |
| | | | 0 | -1246 | 623 | 2 | | | | | | | | | |
| | | | | 0 | -622 | 311 | 2 | | | | | | | | |
| | | | | | 1 | -310 | 155 | 2 | | | | | | | |
| | | | | | | 1 | -154 | 77 | 2 | | | | | | |
| | | | | | | | 1 | -76 | 38 | 2 | | | | | |
| | | | | | | | | 1 | -38 | 19 | 2 | | | | |
| | | | | | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
3988010 = 10011011110010002
the Final answer: 9BC816 = 10011011110010002