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:
200F16 = 2 0 0 F = 2(=0010) 0(=0000) 0(=0000) F(=1111) = 100000000011112
answer: 200F16 = 100000000011112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+0∙162+0∙161+15∙160 = 2∙4096+0∙256+0∙16+15∙1 = 8192+0+0+15 = 820710
got It: 200F16 =820710
Translate the number 820710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
8207 | 2 | | | | | | | | | | | | | |
-8206 | 4103 | 2 | | | | | | | | | | | | |
1 | -4102 | 2051 | 2 | | | | | | | | | | | |
| 1 | -2050 | 1025 | 2 | | | | | | | | | | |
| | 1 | -1024 | 512 | 2 | | | | | | | | | |
| | | 1 | -512 | 256 | 2 | | | | | | | | |
| | | | 0 | -256 | 128 | 2 | | | | | | | |
| | | | | 0 | -128 | 64 | 2 | | | | | | |
| | | | | | 0 | -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:
820710 = 100000000011112
answer: 200F16 = 100000000011112