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:
EB4A16 = E B 4 A = E(=1110) B(=1011) 4(=0100) A(=1010) = 11101011010010102
answer: EB4A16 = 11101011010010102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
14∙163+11∙162+4∙161+10∙160 = 14∙4096+11∙256+4∙16+10∙1 = 57344+2816+64+10 = 6023410
got It: EB4A16 =6023410
Translate the number 6023410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
60234 | 2 | | | | | | | | | | | | | | | |
-60234 | 30117 | 2 | | | | | | | | | | | | | | |
0 | -30116 | 15058 | 2 | | | | | | | | | | | | | |
| 1 | -15058 | 7529 | 2 | | | | | | | | | | | | |
| | 0 | -7528 | 3764 | 2 | | | | | | | | | | | |
| | | 1 | -3764 | 1882 | 2 | | | | | | | | | | |
| | | | 0 | -1882 | 941 | 2 | | | | | | | | | |
| | | | | 0 | -940 | 470 | 2 | | | | | | | | |
| | | | | | 1 | -470 | 235 | 2 | | | | | | | |
| | | | | | | 0 | -234 | 117 | 2 | | | | | | |
| | | | | | | | 1 | -116 | 58 | 2 | | | | | |
| | | | | | | | | 1 | -58 | 29 | 2 | | | | |
| | | | | | | | | | 0 | -28 | 14 | 2 | | | |
| | | | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
6023410 = 11101011010010102
answer: EB4A16 = 11101011010010102