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:
4∙163+12∙162+10∙161+2∙160 = 4∙4096+12∙256+10∙16+2∙1 = 16384+3072+160+2 = 1961810
got It: 4CA216 =1961810
Translate the number 1961810 в octal like this:
the Integer part of the number is divided by the base of the new number system:
19618 | 8 | | | | |
-19616 | 2452 | 8 | | | |
2 | -2448 | 306 | 8 | | |
| 4 | -304 | 38 | 8 | |
| | 2 | -32 | 4 | |
| | | 6 | | |
|
the result of the conversion was:
1961810 = 462428
answer: 4CA216 = 462428
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
4CA216 = 4 C A 2 = 4(=0100) C(=1100) A(=1010) 2(=0010) = 1001100101000102
answer: 4CA216 = 1001100101000102
let\'s make a direct translation from binary to post-binary like this:
1001100101000102 = 100 110 010 100 010 = 100(=4) 110(=6) 010(=2) 100(=4) 010(=2) = 462428
answer: 4CA216 = 462428