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∙163+4∙162+4∙161+13∙160 = 10∙4096+4∙256+4∙16+13∙1 = 40960+1024+64+13 = 4206110
got It: A44D16 =4206110
Translate the number 4206110 в octal like this:
the Integer part of the number is divided by the base of the new number system:
42061 | 8 | | | | | |
-42056 | 5257 | 8 | | | | |
5 | -5256 | 657 | 8 | | | |
| 1 | -656 | 82 | 8 | | |
| | 1 | -80 | 10 | 8 | |
| | | 2 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4206110 = 1221158
answer: A44D16 = 1221158
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
A44D16 = A 4 4 D = A(=1010) 4(=0100) 4(=0100) D(=1101) = 10100100010011012
answer: A44D16 = 10100100010011012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010100100010011012 = 001 010 010 001 001 101 = 001(=1) 010(=2) 010(=2) 001(=1) 001(=1) 101(=5) = 1221158
answer: A44D16 = 1221158