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:
12∙161+6∙160+10∙16-1+6∙16-2 = 12∙16+6∙1+10∙0.0625+6∙0.00390625 = 192+6+0.625+0.0234375 = 198.648437510
got It: C6.A616 =198.648437510
Translate the number 198.648437510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
198 | 8 | | |
-192 | 24 | 8 | |
6 | -24 | 3 | |
| 0 | | |
 |
the Fractional part of the number is multiplied by the base of the new number system:
 |
0. | 6484375*8 |
5 | .1875*8 |
1 | .5*8 |
4 | .0*8 |
the result of the conversion was:
198.648437510 = 306.5148
the Final answer: C6.A616 = 306.5148
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
C6.A616 = C 6. A 6 = C(=1100) 6(=0110). A(=1010) 6(=0110) = 11000110.10100112
the Final answer: C6.A616 = 11000110.10100112
Fill in the number with missing zeros on the left
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
011000110.1010011002 = 011 000 110. 101 001 100 = 011(=3) 000(=0) 110(=6). 101(=5) 001(=1) 100(=4) = 306.5148
the Final answer: C6.A616 = 306.5148