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:
10∙163+11∙162+15∙161+12∙160 = 10∙4096+11∙256+15∙16+12∙1 = 40960+2816+240+12 = 4402810
got It: ABFC16 =4402810
Translate the number 4402810 в octal like this:
the Integer part of the number is divided by the base of the new number system:
44028 | 8 | | | | | |
-44024 | 5503 | 8 | | | | |
4 | -5496 | 687 | 8 | | | |
| 7 | -680 | 85 | 8 | | |
| | 7 | -80 | 10 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4402810 = 1257748
the Final answer: ABFC16 = 1257748
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
ABFC16 = A B F C = A(=1010) B(=1011) F(=1111) C(=1100) = 10101011111111002
the Final answer: ABFC16 = 10101011111111002
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010101011111111002 = 001 010 101 111 111 100 = 001(=1) 010(=2) 101(=5) 111(=7) 111(=7) 100(=4) = 1257748
the Final answer: ABFC16 = 1257748