This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s make a direct translation from binary to post-binary like this:
1101010110011012 = 110 101 011 001 101 = 110(=6) 101(=5) 011(=3) 001(=1) 101(=5) = 653158
the Final answer: 1101010110011012 = 653158
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙214+1∙213+0∙212+1∙211+0∙210+1∙29+0∙28+1∙27+1∙26+0∙25+0∙24+1∙23+1∙22+0∙21+1∙20 = 1∙16384+1∙8192+0∙4096+1∙2048+0∙1024+1∙512+0∙256+1∙128+1∙64+0∙32+0∙16+1∙8+1∙4+0∙2+1∙1 = 16384+8192+0+2048+0+512+0+128+64+0+0+8+4+0+1 = 2734110
got It: 1101010110011012 =2734110
Translate the number 2734110 в octal like this:
the Integer part of the number is divided by the base of the new number system:
27341 | 8 | | | | |
-27336 | 3417 | 8 | | | |
5 | -3416 | 427 | 8 | | |
| 1 | -424 | 53 | 8 | |
| | 3 | -48 | 6 | |
| | | 5 | | |
|
the result of the conversion was:
2734110 = 653158
the Final answer: 1101010110011012 = 653158