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:
5∙162+15∙161+3∙160 = 5∙256+15∙16+3∙1 = 1280+240+3 = 152310
got It: 5f316 =152310
Translate the number 152310 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1523 | 8 | | | |
-1520 | 190 | 8 | | |
3 | -184 | 23 | 8 | |
| 6 | -16 | 2 | |
| | 7 | | |
|
the result of the conversion was:
152310 = 27638
the Final answer: 5f316 = 27638
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
5f316 = 5 f 3 = 5(=0101) f(=1111) 3(=0011) = 101111100112
the Final answer: 5f316 = 101111100112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0101111100112 = 010 111 110 011 = 010(=2) 111(=7) 110(=6) 011(=3) = 27638
the Final answer: 5f316 = 27638