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∙162+0∙161+15∙160+12∙16-1 = 6∙256+0∙16+15∙1+12∙0.0625 = 1536+0+15+0.75 = 1551.7510
got It: 60F.C16 =1551.7510
Translate the number 1551.7510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1551 | 8 | | | |
-1544 | 193 | 8 | | |
7 | -192 | 24 | 8 | |
| 1 | -24 | 3 | |
| | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 75*8 |
6 | .0*8 |
the result of the conversion was:
1551.7510 = 3017.68
answer: 60F.C16 = 3017.68
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
60F.C16 = 6 0 F. C = 6(=0110) 0(=0000) F(=1111). C(=1100) = 11000001111.112
answer: 60F.C16 = 11000001111.112
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:
011000001111.1102 = 011 000 001 111. 110 = 011(=3) 000(=0) 001(=1) 111(=7). 110(=6) = 3017.68
answer: 60F.C16 = 3017.68