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:
E51216 = E 5 1 2 = E(=1110) 5(=0101) 1(=0001) 2(=0010) = 11100101000100102
answer: E51216 = 11100101000100102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
14∙163+5∙162+1∙161+2∙160 = 14∙4096+5∙256+1∙16+2∙1 = 57344+1280+16+2 = 5864210
got It: E51216 =5864210
Translate the number 5864210 в binary like this:
the Integer part of the number is divided by the base of the new number system:
58642 | 2 | | | | | | | | | | | | | | | |
-58642 | 29321 | 2 | | | | | | | | | | | | | | |
0 | -29320 | 14660 | 2 | | | | | | | | | | | | | |
| 1 | -14660 | 7330 | 2 | | | | | | | | | | | | |
| | 0 | -7330 | 3665 | 2 | | | | | | | | | | | |
| | | 0 | -3664 | 1832 | 2 | | | | | | | | | | |
| | | | 1 | -1832 | 916 | 2 | | | | | | | | | |
| | | | | 0 | -916 | 458 | 2 | | | | | | | | |
| | | | | | 0 | -458 | 229 | 2 | | | | | | | |
| | | | | | | 0 | -228 | 114 | 2 | | | | | | |
| | | | | | | | 1 | -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:
5864210 = 11100101000100102
answer: E51216 = 11100101000100102