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:
0f424016 = 0 f 4 2 4 0 = 0(=0000) f(=1111) 4(=0100) 2(=0010) 4(=0100) 0(=0000) = 111101000010010000002
answer: 0f424016 = 111101000010010000002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙165+15∙164+4∙163+2∙162+4∙161+0∙160 = 0∙1048576+15∙65536+4∙4096+2∙256+4∙16+0∙1 = 0+983040+16384+512+64+0 = 100000010
got It: 0f424016 =100000010
Translate the number 100000010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
1000000 | 2 | | | | | | | | | | | | | | | | | | | |
-1000000 | 500000 | 2 | | | | | | | | | | | | | | | | | | |
0 | -500000 | 250000 | 2 | | | | | | | | | | | | | | | | | |
| 0 | -250000 | 125000 | 2 | | | | | | | | | | | | | | | | |
| | 0 | -125000 | 62500 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -62500 | 31250 | 2 | | | | | | | | | | | | | | |
| | | | 0 | -31250 | 15625 | 2 | | | | | | | | | | | | | |
| | | | | 0 | -15624 | 7812 | 2 | | | | | | | | | | | | |
| | | | | | 1 | -7812 | 3906 | 2 | | | | | | | | | | | |
| | | | | | | 0 | -3906 | 1953 | 2 | | | | | | | | | | |
| | | | | | | | 0 | -1952 | 976 | 2 | | | | | | | | | |
| | | | | | | | | 1 | -976 | 488 | 2 | | | | | | | | |
| | | | | | | | | | 0 | -488 | 244 | 2 | | | | | | | |
| | | | | | | | | | | 0 | -244 | 122 | 2 | | | | | | |
| | | | | | | | | | | | 0 | -122 | 61 | 2 | | | | | |
| | | | | | | | | | | | | 0 | -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:
100000010 = 111101000010010000002
answer: 0f424016 = 111101000010010000002