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:
7∙162+12∙161+11∙160 = 7∙256+12∙16+11∙1 = 1792+192+11 = 199510
got It: 7CB16 =199510
Translate the number 199510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1995 | 8 | | | |
-1992 | 249 | 8 | | |
3 | -248 | 31 | 8 | |
| 1 | -24 | 3 | |
| | 7 | | |
|
the result of the conversion was:
199510 = 37138
answer: 7CB16 = 37138
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
7CB16 = 7 C B = 7(=0111) C(=1100) B(=1011) = 111110010112
answer: 7CB16 = 111110010112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0111110010112 = 011 111 001 011 = 011(=3) 111(=7) 001(=1) 011(=3) = 37138
answer: 7CB16 = 37138