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∙163+15∙162+9∙161+1∙160 = 1∙4096+15∙256+9∙16+1∙1 = 4096+3840+144+1 = 808110
got It: 1f9116 =808110
Translate the number 808110 в octal like this:
the Integer part of the number is divided by the base of the new number system:
8081 | 8 | | | | |
-8080 | 1010 | 8 | | | |
1 | -1008 | 126 | 8 | | |
| 2 | -120 | 15 | 8 | |
| | 6 | -8 | 1 | |
| | | 7 | | |
|
the result of the conversion was:
808110 = 176218
answer: 1f9116 = 176218
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
1f9116 = 1 f 9 1 = 1(=0001) f(=1111) 9(=1001) 1(=0001) = 11111100100012
answer: 1f9116 = 11111100100012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011111100100012 = 001 111 110 010 001 = 001(=1) 111(=7) 110(=6) 010(=2) 001(=1) = 176218
answer: 1f9116 = 176218