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:
64CD16 = 6 4 C D = 6(=0110) 4(=0100) C(=1100) D(=1101) = 1100100110011012
the Final answer: 64CD16 = 1100100110011012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
6∙163+4∙162+12∙161+13∙160 = 6∙4096+4∙256+12∙16+13∙1 = 24576+1024+192+13 = 2580510
got It: 64CD16 =2580510
Translate the number 2580510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
25805 | 2 | | | | | | | | | | | | | | |
-25804 | 12902 | 2 | | | | | | | | | | | | | |
1 | -12902 | 6451 | 2 | | | | | | | | | | | | |
| 0 | -6450 | 3225 | 2 | | | | | | | | | | | |
| | 1 | -3224 | 1612 | 2 | | | | | | | | | | |
| | | 1 | -1612 | 806 | 2 | | | | | | | | | |
| | | | 0 | -806 | 403 | 2 | | | | | | | | |
| | | | | 0 | -402 | 201 | 2 | | | | | | | |
| | | | | | 1 | -200 | 100 | 2 | | | | | | |
| | | | | | | 1 | -100 | 50 | 2 | | | | | |
| | | | | | | | 0 | -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:
2580510 = 1100100110011012
the Final answer: 64CD16 = 1100100110011012