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:
53678 = 5 3 6 7 = 5(=101) 3(=011) 6(=110) 7(=111) = 1010111101112
answer: 53678 = 1010111101112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙83+3∙82+6∙81+7∙80 = 5∙512+3∙64+6∙8+7∙1 = 2560+192+48+7 = 280710
got It: 53678 =280710
Translate the number 280710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2807 | 2 | | | | | | | | | | | |
-2806 | 1403 | 2 | | | | | | | | | | |
1 | -1402 | 701 | 2 | | | | | | | | | |
| 1 | -700 | 350 | 2 | | | | | | | | |
| | 1 | -350 | 175 | 2 | | | | | | | |
| | | 0 | -174 | 87 | 2 | | | | | | |
| | | | 1 | -86 | 43 | 2 | | | | | |
| | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | 0 | | |
|
the result of the conversion was:
280710 = 1010111101112
answer: 53678 = 1010111101112