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:
010111018 = 0 1 0 1 1 1 0 1 = 0(=000) 1(=001) 0(=000) 1(=001) 1(=001) 1(=001) 0(=000) 1(=001) = 0000010000010010010000012
answer: 010111018 = 10000010010010000012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙87+1∙86+0∙85+1∙84+1∙83+1∙82+0∙81+1∙80 = 0∙2097152+1∙262144+0∙32768+1∙4096+1∙512+1∙64+0∙8+1∙1 = 0+262144+0+4096+512+64+0+1 = 26681710
got It: 010111018 =26681710
Translate the number 26681710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
266817 | 2 | | | | | | | | | | | | | | | | | | |
-266816 | 133408 | 2 | | | | | | | | | | | | | | | | | |
1 | -133408 | 66704 | 2 | | | | | | | | | | | | | | | | |
| 0 | -66704 | 33352 | 2 | | | | | | | | | | | | | | | |
| | 0 | -33352 | 16676 | 2 | | | | | | | | | | | | | | |
| | | 0 | -16676 | 8338 | 2 | | | | | | | | | | | | | |
| | | | 0 | -8338 | 4169 | 2 | | | | | | | | | | | | |
| | | | | 0 | -4168 | 2084 | 2 | | | | | | | | | | | |
| | | | | | 1 | -2084 | 1042 | 2 | | | | | | | | | | |
| | | | | | | 0 | -1042 | 521 | 2 | | | | | | | | | |
| | | | | | | | 0 | -520 | 260 | 2 | | | | | | | | |
| | | | | | | | | 1 | -260 | 130 | 2 | | | | | | | |
| | | | | | | | | | 0 | -130 | 65 | 2 | | | | | | |
| | | | | | | | | | | 0 | -64 | 32 | 2 | | | | | |
| | | | | | | | | | | | 1 | -32 | 16 | 2 | | | | |
| | | | | | | | | | | | | 0 | -16 | 8 | 2 | | | |
| | | | | | | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
26681710 = 10000010010010000012
answer: 010111018 = 10000010010010000012