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:
2∙162+8∙161+11∙160 = 2∙256+8∙16+11∙1 = 512+128+11 = 65110
got It: 28B16 =65110
Translate the number 65110 в octal like this:
the Integer part of the number is divided by the base of the new number system:
651 | 8 | | | |
-648 | 81 | 8 | | |
3 | -80 | 10 | 8 | |
| 1 | -8 | 1 | |
| | 2 | | |
|
the result of the conversion was:
65110 = 12138
answer: 28B16 = 12138
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
28B16 = 2 8 B = 2(=0010) 8(=1000) B(=1011) = 10100010112
answer: 28B16 = 10100010112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010100010112 = 001 010 001 011 = 001(=1) 010(=2) 001(=1) 011(=3) = 12138
answer: 28B16 = 12138