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 left
let\'s make a direct translation from binary to post-binary like this:
0110010111012 = 011 001 011 101 = 011(=3) 001(=1) 011(=3) 101(=5) = 31358
answer: 110010111012 = 31358
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙211+1∙210+1∙29+0∙28+0∙27+1∙26+0∙25+1∙24+1∙23+1∙22+0∙21+1∙20 = 0∙2048+1∙1024+1∙512+0∙256+0∙128+1∙64+0∙32+1∙16+1∙8+1∙4+0∙2+1∙1 = 0+1024+512+0+0+64+0+16+8+4+0+1 = 162910
got It: 0110010111012 =162910
Translate the number 162910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1629 | 8 | | | |
-1624 | 203 | 8 | | |
5 | -200 | 25 | 8 | |
| 3 | -24 | 3 | |
| | 1 | | |
|
the result of the conversion was:
162910 = 31358
answer: 110010111012 = 31358