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∙161+8∙160+11∙16-1+14∙16-2 = 6∙16+8∙1+11∙0.0625+14∙0.00390625 = 96+8+0.6875+0.0546875 = 104.742187510
got It: 68.BE16 =104.742187510
Translate the number 104.742187510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
104 | 8 | | |
-104 | 13 | 8 | |
0 | -8 | 1 | |
| 5 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 7421875*8 |
5 | .9375*8 |
7 | .5*8 |
4 | .0*8 |
the result of the conversion was:
104.742187510 = 150.5748
answer: 68.BE16 = 150.5748
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
68.BE16 = 6 8. B E = 6(=0110) 8(=1000). B(=1011) E(=1110) = 1101000.10111112
answer: 68.BE16 = 1101000.10111112
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:
001101000.1011111002 = 001 101 000. 101 111 100 = 001(=1) 101(=5) 000(=0). 101(=5) 111(=7) 100(=4) = 150.5748
answer: 68.BE16 = 150.5748