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:
BC116 = B C 1 = B(=1011) C(=1100) 1(=0001) = 1011110000012
answer: BC116 = 1011110000012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙162+12∙161+1∙160 = 11∙256+12∙16+1∙1 = 2816+192+1 = 300910
got It: BC116 =300910
Translate the number 300910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3009 | 2 | | | | | | | | | | | |
-3008 | 1504 | 2 | | | | | | | | | | |
1 | -1504 | 752 | 2 | | | | | | | | | |
| 0 | -752 | 376 | 2 | | | | | | | | |
| | 0 | -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:
300910 = 1011110000012
answer: BC116 = 1011110000012