This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
110001101.1100100111002 = 110 001 101. 110 010 011 100 = 110(=6) 001(=1) 101(=5). 110(=6) 010(=2) 011(=3) 100(=4) = 615.62348
the Final answer: 110001101.11001001112 = 615.62348
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙28+1∙27+0∙26+0∙25+0∙24+1∙23+1∙22+0∙21+1∙20+1∙2-1+1∙2-2+0∙2-3+0∙2-4+1∙2-5+0∙2-6+0∙2-7+1∙2-8+1∙2-9+1∙2-10+0∙2-11+0∙2-12 = 1∙256+1∙128+0∙64+0∙32+0∙16+1∙8+1∙4+0∙2+1∙1+1∙0.5+1∙0.25+0∙0.125+0∙0.0625+1∙0.03125+0∙0.015625+0∙0.0078125+1∙0.00390625+1∙0.001953125+1∙0.0009765625+0∙0.00048828125+0∙0.000244140625 = 256+128+0+0+0+8+4+0+1+0.5+0.25+0+0+0.03125+0+0+0.00390625+0.001953125+0.0009765625+0+0 = 397.788085937510
got It: 110001101.1100100111002 =397.788085937510
Translate the number 397.788085937510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
| 397 | 8 | | |
| -392 | 49 | 8 | |
| 5 | -48 | 6 | |
| 1 | | |
 |
the Fractional part of the number is multiplied by the base of the new number system:
 |
| 0. | 7880859375*8 |
| 6 | .30469*8 |
| 2 | .4375*8 |
| 3 | .5*8 |
| 4 | .0*8 |
the result of the conversion was:
397.788085937510 = 615.62348
the Final answer: 110001101.11001001112 = 615.62348