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:
2F1C16 = 2 F 1 C = 2(=0010) F(=1111) 1(=0001) C(=1100) = 101111000111002
answer: 2F1C16 = 101111000111002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+15∙162+1∙161+12∙160 = 2∙4096+15∙256+1∙16+12∙1 = 8192+3840+16+12 = 1206010
got It: 2F1C16 =1206010
Translate the number 1206010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
12060 | 2 | | | | | | | | | | | | | |
-12060 | 6030 | 2 | | | | | | | | | | | | |
0 | -6030 | 3015 | 2 | | | | | | | | | | | |
| 0 | -3014 | 1507 | 2 | | | | | | | | | | |
| | 1 | -1506 | 753 | 2 | | | | | | | | | |
| | | 1 | -752 | 376 | 2 | | | | | | | | |
| | | | 1 | -376 | 188 | 2 | | | | | | | |
| | | | | 0 | -188 | 94 | 2 | | | | | | |
| | | | | | 0 | -94 | 47 | 2 | | | | | |
| | | | | | | 0 | -46 | 23 | 2 | | | | |
| | | | | | | | 1 | -22 | 11 | 2 | | | |
| | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
1206010 = 101111000111002
answer: 2F1C16 = 101111000111002