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:
AB12316 = A B 1 2 3 = A(=1010) B(=1011) 1(=0001) 2(=0010) 3(=0011) = 101010110001001000112
answer: AB12316 = 101010110001001000112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙164+11∙163+1∙162+2∙161+3∙160 = 10∙65536+11∙4096+1∙256+2∙16+3∙1 = 655360+45056+256+32+3 = 70070710
got It: AB12316 =70070710
Translate the number 70070710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
700707 | 2 | | | | | | | | | | | | | | | | | | | |
-700706 | 350353 | 2 | | | | | | | | | | | | | | | | | | |
1 | -350352 | 175176 | 2 | | | | | | | | | | | | | | | | | |
| 1 | -175176 | 87588 | 2 | | | | | | | | | | | | | | | | |
| | 0 | -87588 | 43794 | 2 | | | | | | | | | | | | | | | |
| | | 0 | -43794 | 21897 | 2 | | | | | | | | | | | | | | |
| | | | 0 | -21896 | 10948 | 2 | | | | | | | | | | | | | |
| | | | | 1 | -10948 | 5474 | 2 | | | | | | | | | | | | |
| | | | | | 0 | -5474 | 2737 | 2 | | | | | | | | | | | |
| | | | | | | 0 | -2736 | 1368 | 2 | | | | | | | | | | |
| | | | | | | | 1 | -1368 | 684 | 2 | | | | | | | | | |
| | | | | | | | | 0 | -684 | 342 | 2 | | | | | | | | |
| | | | | | | | | | 0 | -342 | 171 | 2 | | | | | | | |
| | | | | | | | | | | 0 | -170 | 85 | 2 | | | | | | |
| | | | | | | | | | | | 1 | -84 | 42 | 2 | | | | | |
| | | | | | | | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
70070710 = 101010110001001000112
answer: AB12316 = 101010110001001000112