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∙163+9∙162+10∙161+15∙160 = 2∙4096+9∙256+10∙16+15∙1 = 8192+2304+160+15 = 1067110
got It: 29AF16 =1067110
Translate the number 1067110 в octal like this:
the Integer part of the number is divided by the base of the new number system:
10671 | 8 | | | | |
-10664 | 1333 | 8 | | | |
7 | -1328 | 166 | 8 | | |
| 5 | -160 | 20 | 8 | |
| | 6 | -16 | 2 | |
| | | 4 | | |
|
the result of the conversion was:
1067110 = 246578
answer: 29AF16 = 246578
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
29AF16 = 2 9 A F = 2(=0010) 9(=1001) A(=1010) F(=1111) = 101001101011112
answer: 29AF16 = 101001101011112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0101001101011112 = 010 100 110 101 111 = 010(=2) 100(=4) 110(=6) 101(=5) 111(=7) = 246578
answer: 29AF16 = 246578