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+10∙160 = 2∙16+10∙1 = 32+10 = 4210
got It: 2A16 =4210
Translate the number 4210 в octal like this:
the Integer part of the number is divided by the base of the new number system:
42 | 8 | |
-40 | 5 | |
2 | | |
|
the result of the conversion was:
4210 = 528
answer: 2A16 = 528
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
2A16 = 2 A = 2(=0010) A(=1010) = 1010102
answer: 2A16 = 1010102
let\'s make a direct translation from binary to post-binary like this:
1010102 = 101 010 = 101(=5) 010(=2) = 528
answer: 2A16 = 528