This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from octal to binary like this:
253058 = 2 5 3 0 5 = 2(=010) 5(=101) 3(=011) 0(=000) 5(=101) = 0101010110001012
the Final answer: 253058 = 101010110001012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙84+5∙83+3∙82+0∙81+5∙80 = 2∙4096+5∙512+3∙64+0∙8+5∙1 = 8192+2560+192+0+5 = 1094910
got It: 253058 =1094910
Translate the number 1094910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
10949 | 2 | | | | | | | | | | | | | |
-10948 | 5474 | 2 | | | | | | | | | | | | |
1 | -5474 | 2737 | 2 | | | | | | | | | | | |
| 0 | -2736 | 1368 | 2 | | | | | | | | | | |
| | 1 | -1368 | 684 | 2 | | | | | | | | | |
| | | 0 | -684 | 342 | 2 | | | | | | | | |
| | | | 0 | -342 | 171 | 2 | | | | | | | |
| | | | | 0 | -170 | 85 | 2 | | | | | | |
| | | | | | 1 | -84 | 42 | 2 | | | | | |
| | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
 |
the result of the conversion was:
1094910 = 101010110001012
the Final answer: 253058 = 101010110001012