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:
6∙163+7∙162+10∙161+12∙160 = 6∙4096+7∙256+10∙16+12∙1 = 24576+1792+160+12 = 2654010
got It: 67ac16 =2654010
Translate the number 2654010 в octal like this:
the Integer part of the number is divided by the base of the new number system:
26540 | 8 | | | | |
-26536 | 3317 | 8 | | | |
4 | -3312 | 414 | 8 | | |
| 5 | -408 | 51 | 8 | |
| | 6 | -48 | 6 | |
| | | 3 | | |
|
the result of the conversion was:
2654010 = 636548
answer: 67ac16 = 636548
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
67ac16 = 6 7 a c = 6(=0110) 7(=0111) a(=1010) c(=1100) = 1100111101011002
answer: 67ac16 = 1100111101011002
let\'s make a direct translation from binary to post-binary like this:
1100111101011002 = 110 011 110 101 100 = 110(=6) 011(=3) 110(=6) 101(=5) 100(=4) = 636548
answer: 67ac16 = 636548