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:
BDFF116 = B D F F 1 = B(=1011) D(=1101) F(=1111) F(=1111) 1(=0001) = 101111011111111100012
answer: BDFF116 = 101111011111111100012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
11∙164+13∙163+15∙162+15∙161+1∙160 = 11∙65536+13∙4096+15∙256+15∙16+1∙1 = 720896+53248+3840+240+1 = 77822510
got It: BDFF116 =77822510
Translate the number 77822510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
778225 | 2 | | | | | | | | | | | | | | | | | | | |
-778224 | 389112 | 2 | | | | | | | | | | | | | | | | | | |
1 | -389112 | 194556 | 2 | | | | | | | | | | | | | | | | | |
| 0 | -194556 | 97278 | 2 | | | | | | | | | | | | | | | | |
| | 0 | -97278 | 48639 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -48638 | 24319 | 2 | | | | | | | | | | | | | | |
| | | | 1 | -24318 | 12159 | 2 | | | | | | | | | | | | | |
| | | | | 1 | -12158 | 6079 | 2 | | | | | | | | | | | | |
| | | | | | 1 | -6078 | 3039 | 2 | | | | | | | | | | | |
| | | | | | | 1 | -3038 | 1519 | 2 | | | | | | | | | | |
| | | | | | | | 1 | -1518 | 759 | 2 | | | | | | | | | |
| | | | | | | | | 1 | -758 | 379 | 2 | | | | | | | | |
| | | | | | | | | | 1 | -378 | 189 | 2 | | | | | | | |
| | | | | | | | | | | 1 | -188 | 94 | 2 | | | | | | |
| | | | | | | | | | | | 1 | -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:
77822510 = 101111011111111100012
answer: BDFF116 = 101111011111111100012