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+3∙16-1+13∙16-2 = 2∙16+10∙1+3∙0.0625+13∙0.00390625 = 32+10+0.1875+0.05078125 = 42.2382812510
got It: 2A.3D16 =42.2382812510
Translate the number 42.2382812510 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 Fractional part of the number is multiplied by the base of the new number system:
 |
0. | 23828125*8 |
1 | .90625*8 |
7 | .25*8 |
2 | .0*8 |
the result of the conversion was:
42.2382812510 = 52.1728
the Final answer: 2A.3D16 = 52.1728
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
2A.3D16 = 2 A. 3 D = 2(=0010) A(=1010). 3(=0011) D(=1101) = 101010.001111012
the Final answer: 2A.3D16 = 101010.001111012
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
101010.0011110102 = 101 010. 001 111 010 = 101(=5) 010(=2). 001(=1) 111(=7) 010(=2) = 52.1728
the Final answer: 2A.3D16 = 52.1728