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:
CAD1616 = C A D 1 6 = C(=1100) A(=1010) D(=1101) 1(=0001) 6(=0110) = 110010101101000101102
answer: CAD1616 = 110010101101000101102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙164+10∙163+13∙162+1∙161+6∙160 = 12∙65536+10∙4096+13∙256+1∙16+6∙1 = 786432+40960+3328+16+6 = 83074210
got It: CAD1616 =83074210
Translate the number 83074210 в binary like this:
the Integer part of the number is divided by the base of the new number system:
830742 | 2 | | | | | | | | | | | | | | | | | | | |
-830742 | 415371 | 2 | | | | | | | | | | | | | | | | | | |
0 | -415370 | 207685 | 2 | | | | | | | | | | | | | | | | | |
| 1 | -207684 | 103842 | 2 | | | | | | | | | | | | | | | | |
| | 1 | -103842 | 51921 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -51920 | 25960 | 2 | | | | | | | | | | | | | | |
| | | | 1 | -25960 | 12980 | 2 | | | | | | | | | | | | | |
| | | | | 0 | -12980 | 6490 | 2 | | | | | | | | | | | | |
| | | | | | 0 | -6490 | 3245 | 2 | | | | | | | | | | | |
| | | | | | | 0 | -3244 | 1622 | 2 | | | | | | | | | | |
| | | | | | | | 1 | -1622 | 811 | 2 | | | | | | | | | |
| | | | | | | | | 0 | -810 | 405 | 2 | | | | | | | | |
| | | | | | | | | | 1 | -404 | 202 | 2 | | | | | | | |
| | | | | | | | | | | 1 | -202 | 101 | 2 | | | | | | |
| | | | | | | | | | | | 0 | -100 | 50 | 2 | | | | | |
| | | | | | | | | | | | | 1 | -50 | 25 | 2 | | | | |
| | | | | | | | | | | | | | 0 | -24 | 12 | 2 | | | |
| | | | | | | | | | | | | | | 1 | -12 | 6 | 2 | | |
| | | | | | | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
83074210 = 110010101101000101102
answer: CAD1616 = 110010101101000101102