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:
3E2216 = 3 E 2 2 = 3(=0011) E(=1110) 2(=0010) 2(=0010) = 111110001000102
answer: 3E2216 = 111110001000102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+14∙162+2∙161+2∙160 = 3∙4096+14∙256+2∙16+2∙1 = 12288+3584+32+2 = 1590610
got It: 3E2216 =1590610
Translate the number 1590610 в binary like this:
the Integer part of the number is divided by the base of the new number system:
15906 | 2 | | | | | | | | | | | | | |
-15906 | 7953 | 2 | | | | | | | | | | | | |
0 | -7952 | 3976 | 2 | | | | | | | | | | | |
| 1 | -3976 | 1988 | 2 | | | | | | | | | | |
| | 0 | -1988 | 994 | 2 | | | | | | | | | |
| | | 0 | -994 | 497 | 2 | | | | | | | | |
| | | | 0 | -496 | 248 | 2 | | | | | | | |
| | | | | 1 | -248 | 124 | 2 | | | | | | |
| | | | | | 0 | -124 | 62 | 2 | | | | | |
| | | | | | | 0 | -62 | 31 | 2 | | | | |
| | | | | | | | 0 | -30 | 15 | 2 | | | |
| | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1590610 = 111110001000102
answer: 3E2216 = 111110001000102