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:
10010128 = 1 0 0 1 0 1 2 = 1(=001) 0(=000) 0(=000) 1(=001) 0(=000) 1(=001) 2(=010) = 0010000000010000010102
answer: 10010128 = 10000000010000010102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙86+0∙85+0∙84+1∙83+0∙82+1∙81+2∙80 = 1∙262144+0∙32768+0∙4096+1∙512+0∙64+1∙8+2∙1 = 262144+0+0+512+0+8+2 = 26266610
got It: 10010128 =26266610
Translate the number 26266610 в binary like this:
the Integer part of the number is divided by the base of the new number system:
262666 | 2 | | | | | | | | | | | | | | | | | | |
-262666 | 131333 | 2 | | | | | | | | | | | | | | | | | |
0 | -131332 | 65666 | 2 | | | | | | | | | | | | | | | | |
| 1 | -65666 | 32833 | 2 | | | | | | | | | | | | | | | |
| | 0 | -32832 | 16416 | 2 | | | | | | | | | | | | | | |
| | | 1 | -16416 | 8208 | 2 | | | | | | | | | | | | | |
| | | | 0 | -8208 | 4104 | 2 | | | | | | | | | | | | |
| | | | | 0 | -4104 | 2052 | 2 | | | | | | | | | | | |
| | | | | | 0 | -2052 | 1026 | 2 | | | | | | | | | | |
| | | | | | | 0 | -1026 | 513 | 2 | | | | | | | | | |
| | | | | | | | 0 | -512 | 256 | 2 | | | | | | | | |
| | | | | | | | | 1 | -256 | 128 | 2 | | | | | | | |
| | | | | | | | | | 0 | -128 | 64 | 2 | | | | | | |
| | | | | | | | | | | 0 | -64 | 32 | 2 | | | | | |
| | | | | | | | | | | | 0 | -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:
26266610 = 10000000010000010102
answer: 10010128 = 10000000010000010102