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:
9A3B16 = 9 A 3 B = 9(=1001) A(=1010) 3(=0011) B(=1011) = 10011010001110112
answer: 9A3B16 = 10011010001110112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
9∙163+10∙162+3∙161+11∙160 = 9∙4096+10∙256+3∙16+11∙1 = 36864+2560+48+11 = 3948310
got It: 9A3B16 =3948310
Translate the number 3948310 в binary like this:
the Integer part of the number is divided by the base of the new number system:
39483 | 2 | | | | | | | | | | | | | | | |
-39482 | 19741 | 2 | | | | | | | | | | | | | | |
1 | -19740 | 9870 | 2 | | | | | | | | | | | | | |
| 1 | -9870 | 4935 | 2 | | | | | | | | | | | | |
| | 0 | -4934 | 2467 | 2 | | | | | | | | | | | |
| | | 1 | -2466 | 1233 | 2 | | | | | | | | | | |
| | | | 1 | -1232 | 616 | 2 | | | | | | | | | |
| | | | | 1 | -616 | 308 | 2 | | | | | | | | |
| | | | | | 0 | -308 | 154 | 2 | | | | | | | |
| | | | | | | 0 | -154 | 77 | 2 | | | | | | |
| | | | | | | | 0 | -76 | 38 | 2 | | | | | |
| | | | | | | | | 1 | -38 | 19 | 2 | | | | |
| | | | | | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
3948310 = 10011010001110112
answer: 9A3B16 = 10011010001110112