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:
2∙162+14∙161+13∙160 = 2∙256+14∙16+13∙1 = 512+224+13 = 74910
got It: 2ED16 =74910
Translate the number 74910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
749 | 8 | | | |
-744 | 93 | 8 | | |
5 | -88 | 11 | 8 | |
| 5 | -8 | 1 | |
| | 3 | | |
|
the result of the conversion was:
74910 = 13558
the Final answer: 2ED16 = 13558
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
2ED16 = 2 E D = 2(=0010) E(=1110) D(=1101) = 10111011012
the Final answer: 2ED16 = 10111011012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010111011012 = 001 011 101 101 = 001(=1) 011(=3) 101(=5) 101(=5) = 13558
the Final answer: 2ED16 = 13558