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:
6∙163+14∙162+11∙161+12∙160 = 6∙4096+14∙256+11∙16+12∙1 = 24576+3584+176+12 = 2834810
got It: 6EBC16 =2834810
Translate the number 2834810 в octal like this:
the Integer part of the number is divided by the base of the new number system:
28348 | 8 | | | | |
-28344 | 3543 | 8 | | | |
4 | -3536 | 442 | 8 | | |
| 7 | -440 | 55 | 8 | |
| | 2 | -48 | 6 | |
| | | 7 | | |
|
the result of the conversion was:
2834810 = 672748
the Final answer: 6EBC16 = 672748
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
6EBC16 = 6 E B C = 6(=0110) E(=1110) B(=1011) C(=1100) = 1101110101111002
the Final answer: 6EBC16 = 1101110101111002
let\'s make a direct translation from binary to post-binary like this:
1101110101111002 = 110 111 010 111 100 = 110(=6) 111(=7) 010(=2) 111(=7) 100(=4) = 672748
the Final answer: 6EBC16 = 672748