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:
C4F16 = C 4 F = C(=1100) 4(=0100) F(=1111) = 1100010011112
the Final answer: C4F16 = 1100010011112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙162+4∙161+15∙160 = 12∙256+4∙16+15∙1 = 3072+64+15 = 315110
got It: C4F16 =315110
Translate the number 315110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3151 | 2 | | | | | | | | | | | |
-3150 | 1575 | 2 | | | | | | | | | | |
1 | -1574 | 787 | 2 | | | | | | | | | |
| 1 | -786 | 393 | 2 | | | | | | | | |
| | 1 | -392 | 196 | 2 | | | | | | | |
| | | 1 | -196 | 98 | 2 | | | | | | |
| | | | 0 | -98 | 49 | 2 | | | | | |
| | | | | 0 | -48 | 24 | 2 | | | | |
| | | | | | 1 | -24 | 12 | 2 | | | |
| | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | 1 | | |
|
the result of the conversion was:
315110 = 1100010011112
the Final answer: C4F16 = 1100010011112