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:
10∙162+7∙161+11∙160 = 10∙256+7∙16+11∙1 = 2560+112+11 = 268310
got It: A7B16 =268310
Translate the number 268310 в octal like this:
the Integer part of the number is divided by the base of the new number system:
2683 | 8 | | | |
-2680 | 335 | 8 | | |
3 | -328 | 41 | 8 | |
| 7 | -40 | 5 | |
| | 1 | | |
|
the result of the conversion was:
268310 = 51738
the Final answer: A7B16 = 51738
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
A7B16 = A 7 B = A(=1010) 7(=0111) B(=1011) = 1010011110112
the Final answer: A7B16 = 1010011110112
let\'s make a direct translation from binary to post-binary like this:
1010011110112 = 101 001 111 011 = 101(=5) 001(=1) 111(=7) 011(=3) = 51738
the Final answer: A7B16 = 51738