This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from hexadecimal to binary like this:
6108.4C16 = 6 1 0 8. 4 C = 6(=0110) 1(=0001) 0(=0000) 8(=1000). 4(=0100) C(=1100) = 110000100001000.0100112
the Final answer: 6108.4C16 = 110000100001000.0100112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
6∙163+1∙162+0∙161+8∙160+4∙16-1+12∙16-2 = 6∙4096+1∙256+0∙16+8∙1+4∙0.0625+12∙0.00390625 = 24576+256+0+8+0.25+0.046875 = 24840.29687510
got It: 6108.4C16 =24840.29687510
Translate the number 24840.29687510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
24840 | 2 | | | | | | | | | | | | | | |
-24840 | 12420 | 2 | | | | | | | | | | | | | |
0 | -12420 | 6210 | 2 | | | | | | | | | | | | |
| 0 | -6210 | 3105 | 2 | | | | | | | | | | | |
| | 0 | -3104 | 1552 | 2 | | | | | | | | | | |
| | | 1 | -1552 | 776 | 2 | | | | | | | | | |
| | | | 0 | -776 | 388 | 2 | | | | | | | | |
| | | | | 0 | -388 | 194 | 2 | | | | | | | |
| | | | | | 0 | -194 | 97 | 2 | | | | | | |
| | | | | | | 0 | -96 | 48 | 2 | | | | | |
| | | | | | | | 1 | -48 | 24 | 2 | | | | |
| | | | | | | | | 0 | -24 | 12 | 2 | | | |
| | | | | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | 1 | | |
 |
the Fractional part of the number is multiplied by the base of the new number system:
 |
0. | 296875*2 |
0 | .59375*2 |
1 | .1875*2 |
0 | .375*2 |
0 | .75*2 |
1 | .5*2 |
1 | .0*2 |
the result of the conversion was:
24840.29687510 = 110000100001000.0100112
the Final answer: 6108.4C16 = 110000100001000.0100112