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:
BB9516 = B B 9 5 = B(=1011) B(=1011) 9(=1001) 5(=0101) = 10111011100101012
answer: BB9516 = 10111011100101012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙163+11∙162+9∙161+5∙160 = 11∙4096+11∙256+9∙16+5∙1 = 45056+2816+144+5 = 4802110
got It: BB9516 =4802110
Translate the number 4802110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
48021 | 2 | | | | | | | | | | | | | | | |
-48020 | 24010 | 2 | | | | | | | | | | | | | | |
1 | -24010 | 12005 | 2 | | | | | | | | | | | | | |
| 0 | -12004 | 6002 | 2 | | | | | | | | | | | | |
| | 1 | -6002 | 3001 | 2 | | | | | | | | | | | |
| | | 0 | -3000 | 1500 | 2 | | | | | | | | | | |
| | | | 1 | -1500 | 750 | 2 | | | | | | | | | |
| | | | | 0 | -750 | 375 | 2 | | | | | | | | |
| | | | | | 0 | -374 | 187 | 2 | | | | | | | |
| | | | | | | 1 | -186 | 93 | 2 | | | | | | |
| | | | | | | | 1 | -92 | 46 | 2 | | | | | |
| | | | | | | | | 1 | -46 | 23 | 2 | | | | |
| | | | | | | | | | 0 | -22 | 11 | 2 | | | |
| | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
 |
the result of the conversion was:
4802110 = 10111011100101012
answer: BB9516 = 10111011100101012