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:
FEB116 = F E B 1 = F(=1111) E(=1110) B(=1011) 1(=0001) = 11111110101100012
answer: FEB116 = 11111110101100012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
15∙163+14∙162+11∙161+1∙160 = 15∙4096+14∙256+11∙16+1∙1 = 61440+3584+176+1 = 6520110
got It: FEB116 =6520110
Translate the number 6520110 в binary like this:
the Integer part of the number is divided by the base of the new number system:
65201 | 2 | | | | | | | | | | | | | | | |
-65200 | 32600 | 2 | | | | | | | | | | | | | | |
1 | -32600 | 16300 | 2 | | | | | | | | | | | | | |
| 0 | -16300 | 8150 | 2 | | | | | | | | | | | | |
| | 0 | -8150 | 4075 | 2 | | | | | | | | | | | |
| | | 0 | -4074 | 2037 | 2 | | | | | | | | | | |
| | | | 1 | -2036 | 1018 | 2 | | | | | | | | | |
| | | | | 1 | -1018 | 509 | 2 | | | | | | | | |
| | | | | | 0 | -508 | 254 | 2 | | | | | | | |
| | | | | | | 1 | -254 | 127 | 2 | | | | | | |
| | | | | | | | 0 | -126 | 63 | 2 | | | | | |
| | | | | | | | | 1 | -62 | 31 | 2 | | | | |
| | | | | | | | | | 1 | -30 | 15 | 2 | | | |
| | | | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
6520110 = 11111110101100012
answer: FEB116 = 11111110101100012