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:
10563278 = 1 0 5 6 3 2 7 = 1(=001) 0(=000) 5(=101) 6(=110) 3(=011) 2(=010) 7(=111) = 0010001011100110101112
answer: 10563278 = 10001011100110101112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙86+0∙85+5∙84+6∙83+3∙82+2∙81+7∙80 = 1∙262144+0∙32768+5∙4096+6∙512+3∙64+2∙8+7∙1 = 262144+0+20480+3072+192+16+7 = 28591110
got It: 10563278 =28591110
Translate the number 28591110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
285911 | 2 | | | | | | | | | | | | | | | | | | |
-285910 | 142955 | 2 | | | | | | | | | | | | | | | | | |
1 | -142954 | 71477 | 2 | | | | | | | | | | | | | | | | |
| 1 | -71476 | 35738 | 2 | | | | | | | | | | | | | | | |
| | 1 | -35738 | 17869 | 2 | | | | | | | | | | | | | | |
| | | 0 | -17868 | 8934 | 2 | | | | | | | | | | | | | |
| | | | 1 | -8934 | 4467 | 2 | | | | | | | | | | | | |
| | | | | 0 | -4466 | 2233 | 2 | | | | | | | | | | | |
| | | | | | 1 | -2232 | 1116 | 2 | | | | | | | | | | |
| | | | | | | 1 | -1116 | 558 | 2 | | | | | | | | | |
| | | | | | | | 0 | -558 | 279 | 2 | | | | | | | | |
| | | | | | | | | 0 | -278 | 139 | 2 | | | | | | | |
| | | | | | | | | | 1 | -138 | 69 | 2 | | | | | | |
| | | | | | | | | | | 1 | -68 | 34 | 2 | | | | | |
| | | | | | | | | | | | 1 | -34 | 17 | 2 | | | | |
| | | | | | | | | | | | | 0 | -16 | 8 | 2 | | | |
| | | | | | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
28591110 = 10001011100110101112
answer: 10563278 = 10001011100110101112