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:
8∙163+11∙162+6∙161+14∙160 = 8∙4096+11∙256+6∙16+14∙1 = 32768+2816+96+14 = 3569410
got It: 8B6E16 =3569410
Translate the number 3569410 в octal like this:
the Integer part of the number is divided by the base of the new number system:
35694 | 8 | | | | | |
-35688 | 4461 | 8 | | | | |
6 | -4456 | 557 | 8 | | | |
| 5 | -552 | 69 | 8 | | |
| | 5 | -64 | 8 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 0 | | |
|
the result of the conversion was:
3569410 = 1055568
answer: 8B6E16 = 1055568
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
8B6E16 = 8 B 6 E = 8(=1000) B(=1011) 6(=0110) E(=1110) = 10001011011011102
answer: 8B6E16 = 10001011011011102
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010001011011011102 = 001 000 101 101 101 110 = 001(=1) 000(=0) 101(=5) 101(=5) 101(=5) 110(=6) = 1055568
answer: 8B6E16 = 1055568