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+7∙160 = 4∙256+3∙16+7∙1 = 1024+48+7 = 107910
got It: 43716 =107910
Translate the number 107910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1079 | 8 | | | |
-1072 | 134 | 8 | | |
7 | -128 | 16 | 8 | |
| 6 | -16 | 2 | |
| | 0 | | |
|
the result of the conversion was:
107910 = 20678
answer: 43716 = 20678
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
43716 = 4 3 7 = 4(=0100) 3(=0011) 7(=0111) = 100001101112
answer: 43716 = 100001101112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0100001101112 = 010 000 110 111 = 010(=2) 000(=0) 110(=6) 111(=7) = 20678
answer: 43716 = 20678