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:
493E016 = 4 9 3 E 0 = 4(=0100) 9(=1001) 3(=0011) E(=1110) 0(=0000) = 10010010011111000002
answer: 493E016 = 10010010011111000002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
4∙164+9∙163+3∙162+14∙161+0∙160 = 4∙65536+9∙4096+3∙256+14∙16+0∙1 = 262144+36864+768+224+0 = 30000010
got It: 493E016 =30000010
Translate the number 30000010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
300000 | 2 | | | | | | | | | | | | | | | | | | |
-300000 | 150000 | 2 | | | | | | | | | | | | | | | | | |
0 | -150000 | 75000 | 2 | | | | | | | | | | | | | | | | |
| 0 | -75000 | 37500 | 2 | | | | | | | | | | | | | | | |
| | 0 | -37500 | 18750 | 2 | | | | | | | | | | | | | | |
| | | 0 | -18750 | 9375 | 2 | | | | | | | | | | | | | |
| | | | 0 | -9374 | 4687 | 2 | | | | | | | | | | | | |
| | | | | 1 | -4686 | 2343 | 2 | | | | | | | | | | | |
| | | | | | 1 | -2342 | 1171 | 2 | | | | | | | | | | |
| | | | | | | 1 | -1170 | 585 | 2 | | | | | | | | | |
| | | | | | | | 1 | -584 | 292 | 2 | | | | | | | | |
| | | | | | | | | 1 | -292 | 146 | 2 | | | | | | | |
| | | | | | | | | | 0 | -146 | 73 | 2 | | | | | | |
| | | | | | | | | | | 0 | -72 | 36 | 2 | | | | | |
| | | | | | | | | | | | 1 | -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:
30000010 = 10010010011111000002
answer: 493E016 = 10010010011111000002