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:
11∙163+13∙162+1∙161+15∙160 = 11∙4096+13∙256+1∙16+15∙1 = 45056+3328+16+15 = 4841510
got It: BD1F16 =4841510
Translate the number 4841510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
48415 | 8 | | | | | |
-48408 | 6051 | 8 | | | | |
7 | -6048 | 756 | 8 | | | |
| 3 | -752 | 94 | 8 | | |
| | 4 | -88 | 11 | 8 | |
| | | 6 | -8 | 1 | |
| | | | 3 | | |
 |
the result of the conversion was:
4841510 = 1364378
the Final answer: BD1F16 = 1364378
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
BD1F16 = B D 1 F = B(=1011) D(=1101) 1(=0001) F(=1111) = 10111101000111112
the Final answer: BD1F16 = 10111101000111112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010111101000111112 = 001 011 110 100 011 111 = 001(=1) 011(=3) 110(=6) 100(=4) 011(=3) 111(=7) = 1364378
the Final answer: BD1F16 = 1364378