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:
E48716 = E 4 8 7 = E(=1110) 4(=0100) 8(=1000) 7(=0111) = 11100100100001112
answer: E48716 = 11100100100001112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
14∙163+4∙162+8∙161+7∙160 = 14∙4096+4∙256+8∙16+7∙1 = 57344+1024+128+7 = 5850310
got It: E48716 =5850310
Translate the number 5850310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
58503 | 2 | | | | | | | | | | | | | | | |
-58502 | 29251 | 2 | | | | | | | | | | | | | | |
1 | -29250 | 14625 | 2 | | | | | | | | | | | | | |
| 1 | -14624 | 7312 | 2 | | | | | | | | | | | | |
| | 1 | -7312 | 3656 | 2 | | | | | | | | | | | |
| | | 0 | -3656 | 1828 | 2 | | | | | | | | | | |
| | | | 0 | -1828 | 914 | 2 | | | | | | | | | |
| | | | | 0 | -914 | 457 | 2 | | | | | | | | |
| | | | | | 0 | -456 | 228 | 2 | | | | | | | |
| | | | | | | 1 | -228 | 114 | 2 | | | | | | |
| | | | | | | | 0 | -114 | 57 | 2 | | | | | |
| | | | | | | | | 0 | -56 | 28 | 2 | | | | |
| | | | | | | | | | 1 | -28 | 14 | 2 | | | |
| | | | | | | | | | | 0 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
5850310 = 11100100100001112
answer: E48716 = 11100100100001112