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:
01101018 = 0 1 1 0 1 0 1 = 0(=000) 1(=001) 1(=001) 0(=000) 1(=001) 0(=000) 1(=001) = 0000010010000010000012
answer: 01101018 = 10010000010000012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙86+1∙85+1∙84+0∙83+1∙82+0∙81+1∙80 = 0∙262144+1∙32768+1∙4096+0∙512+1∙64+0∙8+1∙1 = 0+32768+4096+0+64+0+1 = 3692910
got It: 01101018 =3692910
Translate the number 3692910 в binary like this:
the Integer part of the number is divided by the base of the new number system:
36929 | 2 | | | | | | | | | | | | | | | |
-36928 | 18464 | 2 | | | | | | | | | | | | | | |
1 | -18464 | 9232 | 2 | | | | | | | | | | | | | |
| 0 | -9232 | 4616 | 2 | | | | | | | | | | | | |
| | 0 | -4616 | 2308 | 2 | | | | | | | | | | | |
| | | 0 | -2308 | 1154 | 2 | | | | | | | | | | |
| | | | 0 | -1154 | 577 | 2 | | | | | | | | | |
| | | | | 0 | -576 | 288 | 2 | | | | | | | | |
| | | | | | 1 | -288 | 144 | 2 | | | | | | | |
| | | | | | | 0 | -144 | 72 | 2 | | | | | | |
| | | | | | | | 0 | -72 | 36 | 2 | | | | | |
| | | | | | | | | 0 | -36 | 18 | 2 | | | | |
| | | | | | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
3692910 = 10010000010000012
answer: 01101018 = 10010000010000012