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:
14∙161+15∙160 = 14∙16+15∙1 = 224+15 = 23910
got It: ef16 =23910
Translate the number 23910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
239 | 8 | | |
-232 | 29 | 8 | |
7 | -24 | 3 | |
| 5 | | |
|
the result of the conversion was:
23910 = 3578
answer: ef16 = 3578
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
ef16 = e f = e(=1110) f(=1111) = 111011112
answer: ef16 = 111011112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0111011112 = 011 101 111 = 011(=3) 101(=5) 111(=7) = 3578
answer: ef16 = 3578