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:
C4716 = C 4 7 = C(=1100) 4(=0100) 7(=0111) = 1100010001112
answer: C4716 = 1100010001112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙162+4∙161+7∙160 = 12∙256+4∙16+7∙1 = 3072+64+7 = 314310
got It: C4716 =314310
Translate the number 314310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3143 | 2 | | | | | | | | | | | |
-3142 | 1571 | 2 | | | | | | | | | | |
1 | -1570 | 785 | 2 | | | | | | | | | |
| 1 | -784 | 392 | 2 | | | | | | | | |
| | 1 | -392 | 196 | 2 | | | | | | | |
| | | 0 | -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:
314310 = 1100010001112
answer: C4716 = 1100010001112