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:
2∙163+7∙162+4∙161+8∙160 = 2∙4096+7∙256+4∙16+8∙1 = 8192+1792+64+8 = 1005610
got It: 274816 =1005610
Translate the number 1005610 в octal like this:
the Integer part of the number is divided by the base of the new number system:
10056 | 8 | | | | |
-10056 | 1257 | 8 | | | |
0 | -1256 | 157 | 8 | | |
| 1 | -152 | 19 | 8 | |
| | 5 | -16 | 2 | |
| | | 3 | | |
|
the result of the conversion was:
1005610 = 235108
answer: 274816 = 235108
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
274816 = 2 7 4 8 = 2(=0010) 7(=0111) 4(=0100) 8(=1000) = 100111010010002
answer: 274816 = 100111010010002
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0100111010010002 = 010 011 101 001 000 = 010(=2) 011(=3) 101(=5) 001(=1) 000(=0) = 235108
answer: 274816 = 235108