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:
E90716 = E 9 0 7 = E(=1110) 9(=1001) 0(=0000) 7(=0111) = 11101001000001112
the Final answer: E90716 = 11101001000001112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
14∙163+9∙162+0∙161+7∙160 = 14∙4096+9∙256+0∙16+7∙1 = 57344+2304+0+7 = 5965510
got It: E90716 =5965510
Translate the number 5965510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
59655 | 2 | | | | | | | | | | | | | | | |
-59654 | 29827 | 2 | | | | | | | | | | | | | | |
1 | -29826 | 14913 | 2 | | | | | | | | | | | | | |
| 1 | -14912 | 7456 | 2 | | | | | | | | | | | | |
| | 1 | -7456 | 3728 | 2 | | | | | | | | | | | |
| | | 0 | -3728 | 1864 | 2 | | | | | | | | | | |
| | | | 0 | -1864 | 932 | 2 | | | | | | | | | |
| | | | | 0 | -932 | 466 | 2 | | | | | | | | |
| | | | | | 0 | -466 | 233 | 2 | | | | | | | |
| | | | | | | 0 | -232 | 116 | 2 | | | | | | |
| | | | | | | | 1 | -116 | 58 | 2 | | | | | |
| | | | | | | | | 0 | -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:
5965510 = 11101001000001112
the Final answer: E90716 = 11101001000001112