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:
A2BD16 = A 2 B D = A(=1010) 2(=0010) B(=1011) D(=1101) = 10100010101111012
answer: A2BD16 = 10100010101111012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+2∙162+11∙161+13∙160 = 10∙4096+2∙256+11∙16+13∙1 = 40960+512+176+13 = 4166110
got It: A2BD16 =4166110
Translate the number 4166110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
41661 | 2 | | | | | | | | | | | | | | | |
-41660 | 20830 | 2 | | | | | | | | | | | | | | |
1 | -20830 | 10415 | 2 | | | | | | | | | | | | | |
| 0 | -10414 | 5207 | 2 | | | | | | | | | | | | |
| | 1 | -5206 | 2603 | 2 | | | | | | | | | | | |
| | | 1 | -2602 | 1301 | 2 | | | | | | | | | | |
| | | | 1 | -1300 | 650 | 2 | | | | | | | | | |
| | | | | 1 | -650 | 325 | 2 | | | | | | | | |
| | | | | | 0 | -324 | 162 | 2 | | | | | | | |
| | | | | | | 1 | -162 | 81 | 2 | | | | | | |
| | | | | | | | 0 | -80 | 40 | 2 | | | | | |
| | | | | | | | | 1 | -40 | 20 | 2 | | | | |
| | | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4166110 = 10100010101111012
answer: A2BD16 = 10100010101111012