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:
0c03016 = 0 c 0 3 0 = 0(=0000) c(=1100) 0(=0000) 3(=0011) 0(=0000) = 11000000001100002
answer: 0c03016 = 11000000001100002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙164+12∙163+0∙162+3∙161+0∙160 = 0∙65536+12∙4096+0∙256+3∙16+0∙1 = 0+49152+0+48+0 = 4920010
got It: 0c03016 =4920010
Translate the number 4920010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
49200 | 2 | | | | | | | | | | | | | | | |
-49200 | 24600 | 2 | | | | | | | | | | | | | | |
0 | -24600 | 12300 | 2 | | | | | | | | | | | | | |
| 0 | -12300 | 6150 | 2 | | | | | | | | | | | | |
| | 0 | -6150 | 3075 | 2 | | | | | | | | | | | |
| | | 0 | -3074 | 1537 | 2 | | | | | | | | | | |
| | | | 1 | -1536 | 768 | 2 | | | | | | | | | |
| | | | | 1 | -768 | 384 | 2 | | | | | | | | |
| | | | | | 0 | -384 | 192 | 2 | | | | | | | |
| | | | | | | 0 | -192 | 96 | 2 | | | | | | |
| | | | | | | | 0 | -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:
4920010 = 11000000001100002
answer: 0c03016 = 11000000001100002