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:
3∙162+1∙161+11∙160 = 3∙256+1∙16+11∙1 = 768+16+11 = 79510
got It: 31B16 =79510
Translate the number 79510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
795 | 8 | | | |
-792 | 99 | 8 | | |
3 | -96 | 12 | 8 | |
| 3 | -8 | 1 | |
| | 4 | | |
|
the result of the conversion was:
79510 = 14338
answer: 31B16 = 14338
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
31B16 = 3 1 B = 3(=0011) 1(=0001) B(=1011) = 11000110112
answer: 31B16 = 11000110112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011000110112 = 001 100 011 011 = 001(=1) 100(=4) 011(=3) 011(=3) = 14338
answer: 31B16 = 14338