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:
1∙162+0∙161+10∙160 = 1∙256+0∙16+10∙1 = 256+0+10 = 26610
got It: 10A16 =26610
Translate the number 26610 в octal like this:
the Integer part of the number is divided by the base of the new number system:
266 | 8 | | |
-264 | 33 | 8 | |
2 | -32 | 4 | |
| 1 | | |
|
the result of the conversion was:
26610 = 4128
answer: 10A16 = 4128
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
10A16 = 1 0 A = 1(=0001) 0(=0000) A(=1010) = 1000010102
answer: 10A16 = 1000010102
let\'s make a direct translation from binary to post-binary like this:
1000010102 = 100 001 010 = 100(=4) 001(=1) 010(=2) = 4128
answer: 10A16 = 4128