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:
3A4B16 = 3 A 4 B = 3(=0011) A(=1010) 4(=0100) B(=1011) = 111010010010112
the Final answer: 3A4B16 = 111010010010112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+10∙162+4∙161+11∙160 = 3∙4096+10∙256+4∙16+11∙1 = 12288+2560+64+11 = 1492310
got It: 3A4B16 =1492310
Translate the number 1492310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
14923 | 2 | | | | | | | | | | | | | |
-14922 | 7461 | 2 | | | | | | | | | | | | |
1 | -7460 | 3730 | 2 | | | | | | | | | | | |
| 1 | -3730 | 1865 | 2 | | | | | | | | | | |
| | 0 | -1864 | 932 | 2 | | | | | | | | | |
| | | 1 | -932 | 466 | 2 | | | | | | | | |
| | | | 0 | -466 | 233 | 2 | | | | | | | |
| | | | | 0 | -232 | 116 | 2 | | | | | | |
| | | | | | 1 | -116 | 58 | 2 | | | | | |
| | | | | | | 0 | -58 | 29 | 2 | | | | |
| | | | | | | | 0 | -28 | 14 | 2 | | | |
| | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1492310 = 111010010010112
the Final answer: 3A4B16 = 111010010010112