This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s translate to decimal like this:
2∙82+5∙81+5∙80+2∙8-1+5∙8-2+5∙8-3+.∙8-4+2∙8-5+5∙8-6+5∙8-7+.∙8-8+0∙8-9 = 2∙64+5∙8+5∙1+2∙0.125+5∙0.015625+5∙0.001953125+.∙0.000244140625+2∙3.0517578125E-5+5∙3.814697265625E-6+5∙4.7683715820312E-7+.∙5.9604644775391E-8+0∙7.4505805969238E-9 = 128+40+5+0.25+0.078125+0.009765625+0+6.103515625E-5+1.9073486328125E-5+2.3841857910156E-6+0+0 = 173.3379731178283710
got It: 255.255.255.08 =173.3379731178283710
Translate the number 173.3379731178283710 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
173 | 16 | |
-160 | A | |
D | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 33797311782837*16 |
5 | .40757*16 |
6 | .52112*16 |
8 | .33789*16 |
5 | .40625*16 |
6 | .5*16 |
8 | .0*16 |
2 | .0*16 |
the result of the conversion was:
173.3379731178283710 = AD.568568216
answer: 255.255.255.08 = AD.568568216
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
255.255.255.08 = 2 5 5. 2 5 5 . 2 5 5 . 0 = 2(=010) 5(=101) 5(=101). 2(=010) 5(=101) 5(=101) .(=000) 2(=010) 5(=101) 5(=101) .(=000) 0(=000) = 010101101.0101011010000101011010000002
answer: 255.255.255.08 = 10101101.0101011010000101011012
Fill in the number with missing zeros on the right
let\'s do a direct translation from binary to hexadecimal like this:
10101101.0101011010000101011010002 = 1010 1101. 0101 0110 1000 0101 0110 1000 = 1010(=A) 1101(=D). 0101(=5) 0110(=6) 1000(=8) 0101(=5) 0110(=6) 1000(=8) = AD.56856816
answer: 10101101.0101011010000101011010008 = AD.56856816