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:
500708 = 5 0 0 7 0 = 5(=101) 0(=000) 0(=000) 7(=111) 0(=000) = 1010000001110002
answer: 500708 = 1010000001110002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙84+0∙83+0∙82+7∙81+0∙80 = 5∙4096+0∙512+0∙64+7∙8+0∙1 = 20480+0+0+56+0 = 2053610
got It: 500708 =2053610
Translate the number 2053610 в binary like this:
the Integer part of the number is divided by the base of the new number system:
20536 | 2 | | | | | | | | | | | | | | |
-20536 | 10268 | 2 | | | | | | | | | | | | | |
0 | -10268 | 5134 | 2 | | | | | | | | | | | | |
| 0 | -5134 | 2567 | 2 | | | | | | | | | | | |
| | 0 | -2566 | 1283 | 2 | | | | | | | | | | |
| | | 1 | -1282 | 641 | 2 | | | | | | | | | |
| | | | 1 | -640 | 320 | 2 | | | | | | | | |
| | | | | 1 | -320 | 160 | 2 | | | | | | | |
| | | | | | 0 | -160 | 80 | 2 | | | | | | |
| | | | | | | 0 | -80 | 40 | 2 | | | | | |
| | | | | | | | 0 | -40 | 20 | 2 | | | | |
| | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
2053610 = 1010000001110002
answer: 500708 = 1010000001110002