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:
B7C516 = B 7 C 5 = B(=1011) 7(=0111) C(=1100) 5(=0101) = 10110111110001012
answer: B7C516 = 10110111110001012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙163+7∙162+12∙161+5∙160 = 11∙4096+7∙256+12∙16+5∙1 = 45056+1792+192+5 = 4704510
got It: B7C516 =4704510
Translate the number 4704510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
47045 | 2 | | | | | | | | | | | | | | | |
-47044 | 23522 | 2 | | | | | | | | | | | | | | |
1 | -23522 | 11761 | 2 | | | | | | | | | | | | | |
| 0 | -11760 | 5880 | 2 | | | | | | | | | | | | |
| | 1 | -5880 | 2940 | 2 | | | | | | | | | | | |
| | | 0 | -2940 | 1470 | 2 | | | | | | | | | | |
| | | | 0 | -1470 | 735 | 2 | | | | | | | | | |
| | | | | 0 | -734 | 367 | 2 | | | | | | | | |
| | | | | | 1 | -366 | 183 | 2 | | | | | | | |
| | | | | | | 1 | -182 | 91 | 2 | | | | | | |
| | | | | | | | 1 | -90 | 45 | 2 | | | | | |
| | | | | | | | | 1 | -44 | 22 | 2 | | | | |
| | | | | | | | | | 1 | -22 | 11 | 2 | | | |
| | | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4704510 = 10110111110001012
answer: B7C516 = 10110111110001012