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:
00003DCC16 = 0 0 0 0 3 D C C = 0(=0000) 0(=0000) 0(=0000) 0(=0000) 3(=0011) D(=1101) C(=1100) C(=1100) = 111101110011002
the Final answer: 00003DCC16 = 111101110011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙167+0∙166+0∙165+0∙164+3∙163+13∙162+12∙161+12∙160 = 0∙268435456+0∙16777216+0∙1048576+0∙65536+3∙4096+13∙256+12∙16+12∙1 = 0+0+0+0+12288+3328+192+12 = 1582010
got It: 00003DCC16 =1582010
Translate the number 1582010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
15820 | 2 | | | | | | | | | | | | | |
-15820 | 7910 | 2 | | | | | | | | | | | | |
0 | -7910 | 3955 | 2 | | | | | | | | | | | |
| 0 | -3954 | 1977 | 2 | | | | | | | | | | |
| | 1 | -1976 | 988 | 2 | | | | | | | | | |
| | | 1 | -988 | 494 | 2 | | | | | | | | |
| | | | 0 | -494 | 247 | 2 | | | | | | | |
| | | | | 0 | -246 | 123 | 2 | | | | | | |
| | | | | | 1 | -122 | 61 | 2 | | | | | |
| | | | | | | 1 | -60 | 30 | 2 | | | | |
| | | | | | | | 1 | -30 | 15 | 2 | | | |
| | | | | | | | | 0 | -14 | 7 | 2 | | |
| | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1582010 = 111101110011002
the Final answer: 00003DCC16 = 111101110011002