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:
15C3816 = 1 5 C 3 8 = 1(=0001) 5(=0101) C(=1100) 3(=0011) 8(=1000) = 101011100001110002
answer: 15C3816 = 101011100001110002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙164+5∙163+12∙162+3∙161+8∙160 = 1∙65536+5∙4096+12∙256+3∙16+8∙1 = 65536+20480+3072+48+8 = 8914410
got It: 15C3816 =8914410
Translate the number 8914410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
89144 | 2 | | | | | | | | | | | | | | | | |
-89144 | 44572 | 2 | | | | | | | | | | | | | | | |
0 | -44572 | 22286 | 2 | | | | | | | | | | | | | | |
| 0 | -22286 | 11143 | 2 | | | | | | | | | | | | | |
| | 0 | -11142 | 5571 | 2 | | | | | | | | | | | | |
| | | 1 | -5570 | 2785 | 2 | | | | | | | | | | | |
| | | | 1 | -2784 | 1392 | 2 | | | | | | | | | | |
| | | | | 1 | -1392 | 696 | 2 | | | | | | | | | |
| | | | | | 0 | -696 | 348 | 2 | | | | | | | | |
| | | | | | | 0 | -348 | 174 | 2 | | | | | | | |
| | | | | | | | 0 | -174 | 87 | 2 | | | | | | |
| | | | | | | | | 0 | -86 | 43 | 2 | | | | | |
| | | | | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
8914410 = 101011100001110002
answer: 15C3816 = 101011100001110002