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:
2C6B16 = 2 C 6 B = 2(=0010) C(=1100) 6(=0110) B(=1011) = 101100011010112
answer: 2C6B16 = 101100011010112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+12∙162+6∙161+11∙160 = 2∙4096+12∙256+6∙16+11∙1 = 8192+3072+96+11 = 1137110
got It: 2C6B16 =1137110
Translate the number 1137110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
11371 | 2 | | | | | | | | | | | | | |
-11370 | 5685 | 2 | | | | | | | | | | | | |
1 | -5684 | 2842 | 2 | | | | | | | | | | | |
| 1 | -2842 | 1421 | 2 | | | | | | | | | | |
| | 0 | -1420 | 710 | 2 | | | | | | | | | |
| | | 1 | -710 | 355 | 2 | | | | | | | | |
| | | | 0 | -354 | 177 | 2 | | | | | | | |
| | | | | 1 | -176 | 88 | 2 | | | | | | |
| | | | | | 1 | -88 | 44 | 2 | | | | | |
| | | | | | | 0 | -44 | 22 | 2 | | | | |
| | | | | | | | 0 | -22 | 11 | 2 | | | |
| | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
1137110 = 101100011010112
answer: 2C6B16 = 101100011010112