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∙161+12∙160 = 2∙16+12∙1 = 32+12 = 4410
got It: 2C16 =4410
Translate the number 4410 в octal like this:
the Integer part of the number is divided by the base of the new number system:
44 | 8 | |
-40 | 5 | |
4 | | |
|
the result of the conversion was:
4410 = 548
the Final answer: 2C16 = 548
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
2C16 = 2 C = 2(=0010) C(=1100) = 1011002
the Final answer: 2C16 = 1011002
let\'s make a direct translation from binary to post-binary like this:
1011002 = 101 100 = 101(=5) 100(=4) = 548
the Final answer: 2C16 = 548