This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
12∙163+0∙162+7∙161+9∙160 = 12∙4096+0∙256+7∙16+9∙1 = 49152+0+112+9 = 4927310
got It: C07916 =4927310
Translate the number 4927310 в octal like this:
the Integer part of the number is divided by the base of the new number system:
49273 | 8 | | | | | |
-49272 | 6159 | 8 | | | | |
1 | -6152 | 769 | 8 | | | |
| 7 | -768 | 96 | 8 | | |
| | 1 | -96 | 12 | 8 | |
| | | 0 | -8 | 1 | |
| | | | 4 | | |
|
the result of the conversion was:
4927310 = 1401718
answer: C07916 = 1401718
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
C07916 = C 0 7 9 = C(=1100) 0(=0000) 7(=0111) 9(=1001) = 11000000011110012
answer: C07916 = 11000000011110012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011000000011110012 = 001 100 000 001 111 001 = 001(=1) 100(=4) 000(=0) 001(=1) 111(=7) 001(=1) = 1401718
answer: C07916 = 1401718