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:
9E8B316 = 9 E 8 B 3 = 9(=1001) E(=1110) 8(=1000) B(=1011) 3(=0011) = 100111101000101100112
answer: 9E8B316 = 100111101000101100112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
9∙164+14∙163+8∙162+11∙161+3∙160 = 9∙65536+14∙4096+8∙256+11∙16+3∙1 = 589824+57344+2048+176+3 = 64939510
got It: 9E8B316 =64939510
Translate the number 64939510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
649395 | 2 | | | | | | | | | | | | | | | | | | | |
-649394 | 324697 | 2 | | | | | | | | | | | | | | | | | | |
1 | -324696 | 162348 | 2 | | | | | | | | | | | | | | | | | |
| 1 | -162348 | 81174 | 2 | | | | | | | | | | | | | | | | |
| | 0 | -81174 | 40587 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -40586 | 20293 | 2 | | | | | | | | | | | | | | |
| | | | 1 | -20292 | 10146 | 2 | | | | | | | | | | | | | |
| | | | | 1 | -10146 | 5073 | 2 | | | | | | | | | | | | |
| | | | | | 0 | -5072 | 2536 | 2 | | | | | | | | | | | |
| | | | | | | 1 | -2536 | 1268 | 2 | | | | | | | | | | |
| | | | | | | | 0 | -1268 | 634 | 2 | | | | | | | | | |
| | | | | | | | | 0 | -634 | 317 | 2 | | | | | | | | |
| | | | | | | | | | 0 | -316 | 158 | 2 | | | | | | | |
| | | | | | | | | | | 1 | -158 | 79 | 2 | | | | | | |
| | | | | | | | | | | | 0 | -78 | 39 | 2 | | | | | |
| | | | | | | | | | | | | 1 | -38 | 19 | 2 | | | | |
| | | | | | | | | | | | | | 1 | -18 | 9 | 2 | | | |
| | | | | | | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
64939510 = 100111101000101100112
answer: 9E8B316 = 100111101000101100112