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:
C1816 = C 1 8 = C(=1100) 1(=0001) 8(=1000) = 1100000110002
the Final answer: C1816 = 1100000110002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙162+1∙161+8∙160 = 12∙256+1∙16+8∙1 = 3072+16+8 = 309610
got It: C1816 =309610
Translate the number 309610 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3096 | 2 | | | | | | | | | | | |
-3096 | 1548 | 2 | | | | | | | | | | |
0 | -1548 | 774 | 2 | | | | | | | | | |
| 0 | -774 | 387 | 2 | | | | | | | | |
| | 0 | -386 | 193 | 2 | | | | | | | |
| | | 1 | -192 | 96 | 2 | | | | | | |
| | | | 1 | -96 | 48 | 2 | | | | | |
| | | | | 0 | -48 | 24 | 2 | | | | |
| | | | | | 0 | -24 | 12 | 2 | | | |
| | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | 1 | | |
|
the result of the conversion was:
309610 = 1100000110002
the Final answer: C1816 = 1100000110002