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:
123AB16 = 1 2 3 A B = 1(=0001) 2(=0010) 3(=0011) A(=1010) B(=1011) = 100100011101010112
answer: 123AB16 = 100100011101010112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙164+2∙163+3∙162+10∙161+11∙160 = 1∙65536+2∙4096+3∙256+10∙16+11∙1 = 65536+8192+768+160+11 = 7466710
got It: 123AB16 =7466710
Translate the number 7466710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
74667 | 2 | | | | | | | | | | | | | | | | |
-74666 | 37333 | 2 | | | | | | | | | | | | | | | |
1 | -37332 | 18666 | 2 | | | | | | | | | | | | | | |
| 1 | -18666 | 9333 | 2 | | | | | | | | | | | | | |
| | 0 | -9332 | 4666 | 2 | | | | | | | | | | | | |
| | | 1 | -4666 | 2333 | 2 | | | | | | | | | | | |
| | | | 0 | -2332 | 1166 | 2 | | | | | | | | | | |
| | | | | 1 | -1166 | 583 | 2 | | | | | | | | | |
| | | | | | 0 | -582 | 291 | 2 | | | | | | | | |
| | | | | | | 1 | -290 | 145 | 2 | | | | | | | |
| | | | | | | | 1 | -144 | 72 | 2 | | | | | | |
| | | | | | | | | 1 | -72 | 36 | 2 | | | | | |
| | | | | | | | | | 0 | -36 | 18 | 2 | | | | |
| | | | | | | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
7466710 = 100100011101010112
answer: 123AB16 = 100100011101010112