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:
1∙162+10∙161+12∙160 = 1∙256+10∙16+12∙1 = 256+160+12 = 42810
got It: 1AC16 =42810
Translate the number 42810 в octal like this:
the Integer part of the number is divided by the base of the new number system:
428 | 8 | | |
-424 | 53 | 8 | |
4 | -48 | 6 | |
| 5 | | |
|
the result of the conversion was:
42810 = 6548
answer: 1AC16 = 6548
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
1AC16 = 1 A C = 1(=0001) A(=1010) C(=1100) = 1101011002
answer: 1AC16 = 1101011002
let\'s make a direct translation from binary to post-binary like this:
1101011002 = 110 101 100 = 110(=6) 101(=5) 100(=4) = 6548
answer: 1AC16 = 6548