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+10∙162+11∙161+11∙160 = 10∙4096+10∙256+11∙16+11∙1 = 40960+2560+176+11 = 4370710
got It: AABB16 =4370710
Translate the number 4370710 в octal like this:
the Integer part of the number is divided by the base of the new number system:
43707 | 8 | | | | | |
-43704 | 5463 | 8 | | | | |
3 | -5456 | 682 | 8 | | | |
| 7 | -680 | 85 | 8 | | |
| | 2 | -80 | 10 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4370710 = 1252738
answer: AABB16 = 1252738
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
AABB16 = A A B B = A(=1010) A(=1010) B(=1011) B(=1011) = 10101010101110112
answer: AABB16 = 10101010101110112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010101010101110112 = 001 010 101 010 111 011 = 001(=1) 010(=2) 101(=5) 010(=2) 111(=7) 011(=3) = 1252738
answer: AABB16 = 1252738