This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
13∙163+12∙162+11∙161+10∙160 = 13∙4096+12∙256+11∙16+10∙1 = 53248+3072+176+10 = 5650610
got It: dcba16 =5650610
Translate the number 5650610 в octal like this:
the Integer part of the number is divided by the base of the new number system:
56506 | 8 | | | | | |
-56504 | 7063 | 8 | | | | |
2 | -7056 | 882 | 8 | | | |
| 7 | -880 | 110 | 8 | | |
| | 2 | -104 | 13 | 8 | |
| | | 6 | -8 | 1 | |
| | | | 5 | | |
|
the result of the conversion was:
5650610 = 1562728
answer: dcba16 = 1562728
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
dcba16 = d c b a = d(=1101) c(=1100) b(=1011) a(=1010) = 11011100101110102
answer: dcba16 = 11011100101110102
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011011100101110102 = 001 101 110 010 111 010 = 001(=1) 101(=5) 110(=6) 010(=2) 111(=7) 010(=2) = 1562728
answer: dcba16 = 1562728