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∙161+14∙160+5∙16-1+3∙16-2 = 1∙16+14∙1+5∙0.0625+3∙0.00390625 = 16+14+0.3125+0.01171875 = 30.3242187510
got It: 1E.5316 =30.3242187510
Translate the number 30.3242187510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
30 | 8 | |
-24 | 3 | |
6 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 32421875*8 |
2 | .59375*8 |
4 | .75*8 |
6 | .0*8 |
the result of the conversion was:
30.3242187510 = 36.2468
the Final answer: 1E.5316 = 36.2468
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
1E.5316 = 1 E. 5 3 = 1(=0001) E(=1110). 5(=0101) 3(=0011) = 11110.010100112
the Final answer: 1E.5316 = 11110.010100112
Fill in the number with missing zeros on the left
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
011110.0101001102 = 011 110. 010 100 110 = 011(=3) 110(=6). 010(=2) 100(=4) 110(=6) = 36.2468
the Final answer: 1E.5316 = 36.2468