This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s translate to decimal like this:
7∙84+4∙83+3∙82+2∙81+1∙80 = 7∙4096+4∙512+3∙64+2∙8+1∙1 = 28672+2048+192+16+1 = 3092910
got It: 743218 =3092910
Translate the number 3092910 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
30929 | 16 | | | |
-30928 | 1933 | 16 | | |
1 | -1920 | 120 | 16 | |
| D | -112 | 7 | |
| | 8 | | |
|
the result of the conversion was:
3092910 = 78D116
answer: 743218 = 78D116
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
743218 = 7 4 3 2 1 = 7(=111) 4(=100) 3(=011) 2(=010) 1(=001) = 1111000110100012
answer: 743218 = 1111000110100012
Fill in the number with missing zeros on the left
let\'s do a direct translation from binary to hexadecimal like this:
01111000110100012 = 0111 1000 1101 0001 = 0111(=7) 1000(=8) 1101(=D) 0001(=1) = 78D116
answer: 01111000110100018 = 78D116