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:
bca116 = b c a 1 = b(=1011) c(=1100) a(=1010) 1(=0001) = 10111100101000012
the Final answer: bca116 = 10111100101000012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙163+12∙162+10∙161+1∙160 = 11∙4096+12∙256+10∙16+1∙1 = 45056+3072+160+1 = 4828910
got It: bca116 =4828910
Translate the number 4828910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
48289 | 2 | | | | | | | | | | | | | | | |
-48288 | 24144 | 2 | | | | | | | | | | | | | | |
1 | -24144 | 12072 | 2 | | | | | | | | | | | | | |
| 0 | -12072 | 6036 | 2 | | | | | | | | | | | | |
| | 0 | -6036 | 3018 | 2 | | | | | | | | | | | |
| | | 0 | -3018 | 1509 | 2 | | | | | | | | | | |
| | | | 0 | -1508 | 754 | 2 | | | | | | | | | |
| | | | | 1 | -754 | 377 | 2 | | | | | | | | |
| | | | | | 0 | -376 | 188 | 2 | | | | | | | |
| | | | | | | 1 | -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:
4828910 = 10111100101000012
the Final answer: bca116 = 10111100101000012