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:
15EC16 = 1 5 E C = 1(=0001) 5(=0101) E(=1110) C(=1100) = 10101111011002
answer: 15EC16 = 10101111011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙163+5∙162+14∙161+12∙160 = 1∙4096+5∙256+14∙16+12∙1 = 4096+1280+224+12 = 561210
got It: 15EC16 =561210
Translate the number 561210 в binary like this:
the Integer part of the number is divided by the base of the new number system:
5612 | 2 | | | | | | | | | | | | |
-5612 | 2806 | 2 | | | | | | | | | | | |
0 | -2806 | 1403 | 2 | | | | | | | | | | |
| 0 | -1402 | 701 | 2 | | | | | | | | | |
| | 1 | -700 | 350 | 2 | | | | | | | | |
| | | 1 | -350 | 175 | 2 | | | | | | | |
| | | | 0 | -174 | 87 | 2 | | | | | | |
| | | | | 1 | -86 | 43 | 2 | | | | | |
| | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | 0 | | |
|
the result of the conversion was:
561210 = 10101111011002
answer: 15EC16 = 10101111011002