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∙163+10∙162+15∙161+14∙160+3∙16-1+13∙16-2 = 12∙4096+10∙256+15∙16+14∙1+3∙0.0625+13∙0.00390625 = 49152+2560+240+14+0.1875+0.05078125 = 51966.2382812510
got It: Cafe.3D16 =51966.2382812510
Translate the number 51966.2382812510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
51966 | 8 | | | | | |
-51960 | 6495 | 8 | | | | |
6 | -6488 | 811 | 8 | | | |
| 7 | -808 | 101 | 8 | | |
| | 3 | -96 | 12 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 4 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 23828125*8 |
1 | .90625*8 |
7 | .25*8 |
2 | .0*8 |
the result of the conversion was:
51966.2382812510 = 145376.1728
answer: Cafe.3D16 = 145376.1728
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
Cafe.3D16 = C a f e. 3 D = C(=1100) a(=1010) f(=1111) e(=1110). 3(=0011) D(=1101) = 1100101011111110.001111012
answer: Cafe.3D16 = 1100101011111110.001111012
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:
001100101011111110.0011110102 = 001 100 101 011 111 110. 001 111 010 = 001(=1) 100(=4) 101(=5) 011(=3) 111(=7) 110(=6). 001(=1) 111(=7) 010(=2) = 145376.1728
answer: Cafe.3D16 = 145376.1728