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:
40AE16 = 4 0 A E = 4(=0100) 0(=0000) A(=1010) E(=1110) = 1000000101011102
answer: 40AE16 = 1000000101011102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
4∙163+0∙162+10∙161+14∙160 = 4∙4096+0∙256+10∙16+14∙1 = 16384+0+160+14 = 1655810
got It: 40AE16 =1655810
Translate the number 1655810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
16558 | 2 | | | | | | | | | | | | | | |
-16558 | 8279 | 2 | | | | | | | | | | | | | |
0 | -8278 | 4139 | 2 | | | | | | | | | | | | |
| 1 | -4138 | 2069 | 2 | | | | | | | | | | | |
| | 1 | -2068 | 1034 | 2 | | | | | | | | | | |
| | | 1 | -1034 | 517 | 2 | | | | | | | | | |
| | | | 0 | -516 | 258 | 2 | | | | | | | | |
| | | | | 1 | -258 | 129 | 2 | | | | | | | |
| | | | | | 0 | -128 | 64 | 2 | | | | | | |
| | | | | | | 1 | -64 | 32 | 2 | | | | | |
| | | | | | | | 0 | -32 | 16 | 2 | | | | |
| | | | | | | | | 0 | -16 | 8 | 2 | | | |
| | | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
1655810 = 1000000101011102
answer: 40AE16 = 1000000101011102