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:
4∙162+3∙161+13∙160 = 4∙256+3∙16+13∙1 = 1024+48+13 = 108510
got It: 43D16 =108510
Translate the number 108510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1085 | 8 | | | |
-1080 | 135 | 8 | | |
5 | -128 | 16 | 8 | |
| 7 | -16 | 2 | |
| | 0 | | |
|
the result of the conversion was:
108510 = 20758
answer: 43D16 = 20758
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
43D16 = 4 3 D = 4(=0100) 3(=0011) D(=1101) = 100001111012
answer: 43D16 = 100001111012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0100001111012 = 010 000 111 101 = 010(=2) 000(=0) 111(=7) 101(=5) = 20758
answer: 43D16 = 20758