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:
51548 = 5 1 5 4 = 5(=101) 1(=001) 5(=101) 4(=100) = 1010011011002
answer: 51548 = 1010011011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙83+1∙82+5∙81+4∙80 = 5∙512+1∙64+5∙8+4∙1 = 2560+64+40+4 = 266810
got It: 51548 =266810
Translate the number 266810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2668 | 2 | | | | | | | | | | | |
-2668 | 1334 | 2 | | | | | | | | | | |
0 | -1334 | 667 | 2 | | | | | | | | | |
| 0 | -666 | 333 | 2 | | | | | | | | |
| | 1 | -332 | 166 | 2 | | | | | | | |
| | | 1 | -166 | 83 | 2 | | | | | | |
| | | | 0 | -82 | 41 | 2 | | | | | |
| | | | | 1 | -40 | 20 | 2 | | | | |
| | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | 0 | | |
|
the result of the conversion was:
266810 = 1010011011002
answer: 51548 = 1010011011002