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+1∙161+0∙160 = 10∙4096+11∙256+1∙16+0∙1 = 40960+2816+16+0 = 4379210
got It: AB1016 =4379210
Translate the number 4379210 в octal like this:
the Integer part of the number is divided by the base of the new number system:
43792 | 8 | | | | | |
-43792 | 5474 | 8 | | | | |
0 | -5472 | 684 | 8 | | | |
| 2 | -680 | 85 | 8 | | |
| | 4 | -80 | 10 | 8 | |
| | | 5 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4379210 = 1254208
answer: AB1016 = 1254208
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
AB1016 = A B 1 0 = A(=1010) B(=1011) 1(=0001) 0(=0000) = 10101011000100002
answer: AB1016 = 10101011000100002
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010101011000100002 = 001 010 101 100 010 000 = 001(=1) 010(=2) 101(=5) 100(=4) 010(=2) 000(=0) = 1254208
answer: AB1016 = 1254208