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:
0111111110001002 = 011 111 111 000 100 = 011(=3) 111(=7) 111(=7) 000(=0) 100(=4) = 377048
answer: 111111110001002 = 377048
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙214+1∙213+1∙212+1∙211+1∙210+1∙29+1∙28+1∙27+1∙26+0∙25+0∙24+0∙23+1∙22+0∙21+0∙20 = 0∙16384+1∙8192+1∙4096+1∙2048+1∙1024+1∙512+1∙256+1∙128+1∙64+0∙32+0∙16+0∙8+1∙4+0∙2+0∙1 = 0+8192+4096+2048+1024+512+256+128+64+0+0+0+4+0+0 = 1632410
got It: 0111111110001002 =1632410
Translate the number 1632410 в octal like this:
the Integer part of the number is divided by the base of the new number system:
16324 | 8 | | | | |
-16320 | 2040 | 8 | | | |
4 | -2040 | 255 | 8 | | |
| 0 | -248 | 31 | 8 | |
| | 7 | -24 | 3 | |
| | | 7 | | |
|
the result of the conversion was:
1632410 = 377048
answer: 111111110001002 = 377048