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+15∙160 = 10∙4096+11∙256+14∙16+15∙1 = 40960+2816+224+15 = 4401510
got It: ABEF16 =4401510
Translate the number 4401510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
44015 | 8 | | | | | |
-44008 | 5501 | 8 | | | | |
7 | -5496 | 687 | 8 | | | |
| 5 | -680 | 85 | 8 | | |
| | 7 | -80 | 10 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4401510 = 1257578
answer: ABEF16 = 1257578
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
ABEF16 = A B E F = A(=1010) B(=1011) E(=1110) F(=1111) = 10101011111011112
answer: ABEF16 = 10101011111011112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010101011111011112 = 001 010 101 111 101 111 = 001(=1) 010(=2) 101(=5) 111(=7) 101(=5) 111(=7) = 1257578
answer: ABEF16 = 1257578