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+14∙161+13∙160 = 10∙4096+11∙256+14∙16+13∙1 = 40960+2816+224+13 = 4401310
got It: ABED16 =4401310
Translate the number 4401310 в octal like this:
the Integer part of the number is divided by the base of the new number system:
44013 | 8 | | | | | |
-44008 | 5501 | 8 | | | | |
5 | -5496 | 687 | 8 | | | |
| 5 | -680 | 85 | 8 | | |
| | 7 | -80 | 10 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4401310 = 1257558
answer: ABED16 = 1257558
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
ABED16 = A B E D = A(=1010) B(=1011) E(=1110) D(=1101) = 10101011111011012
answer: ABED16 = 10101011111011012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010101011111011012 = 001 010 101 111 101 101 = 001(=1) 010(=2) 101(=5) 111(=7) 101(=5) 101(=5) = 1257558
answer: ABED16 = 1257558