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+1∙161+3∙160 = 10∙256+1∙16+3∙1 = 2560+16+3 = 257910
got It: A1316 =257910
Translate the number 257910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
2579 | 8 | | | |
-2576 | 322 | 8 | | |
3 | -320 | 40 | 8 | |
| 2 | -40 | 5 | |
| | 0 | | |
|
the result of the conversion was:
257910 = 50238
the Final answer: A1316 = 50238
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
A1316 = A 1 3 = A(=1010) 1(=0001) 3(=0011) = 1010000100112
the Final answer: A1316 = 1010000100112
let\'s make a direct translation from binary to post-binary like this:
1010000100112 = 101 000 010 011 = 101(=5) 000(=0) 010(=2) 011(=3) = 50238
the Final answer: A1316 = 50238