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:
0111101012 = 011 110 101 = 011(=3) 110(=6) 101(=5) = 3658
the Final answer: 111101012 = 3658
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙28+1∙27+1∙26+1∙25+1∙24+0∙23+1∙22+0∙21+1∙20 = 0∙256+1∙128+1∙64+1∙32+1∙16+0∙8+1∙4+0∙2+1∙1 = 0+128+64+32+16+0+4+0+1 = 24510
got It: 0111101012 =24510
Translate the number 24510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
245 | 8 | | |
-240 | 30 | 8 | |
5 | -24 | 3 | |
| 6 | | |
|
the result of the conversion was:
24510 = 3658
the Final answer: 111101012 = 3658