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:
3CE616 = 3 C E 6 = 3(=0011) C(=1100) E(=1110) 6(=0110) = 111100111001102
the Final answer: 3CE616 = 111100111001102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+12∙162+14∙161+6∙160 = 3∙4096+12∙256+14∙16+6∙1 = 12288+3072+224+6 = 1559010
got It: 3CE616 =1559010
Translate the number 1559010 в binary like this:
the Integer part of the number is divided by the base of the new number system:
15590 | 2 | | | | | | | | | | | | | |
-15590 | 7795 | 2 | | | | | | | | | | | | |
0 | -7794 | 3897 | 2 | | | | | | | | | | | |
| 1 | -3896 | 1948 | 2 | | | | | | | | | | |
| | 1 | -1948 | 974 | 2 | | | | | | | | | |
| | | 0 | -974 | 487 | 2 | | | | | | | | |
| | | | 0 | -486 | 243 | 2 | | | | | | | |
| | | | | 1 | -242 | 121 | 2 | | | | | | |
| | | | | | 1 | -120 | 60 | 2 | | | | | |
| | | | | | | 1 | -60 | 30 | 2 | | | | |
| | | | | | | | 0 | -30 | 15 | 2 | | | |
| | | | | | | | | 0 | -14 | 7 | 2 | | |
| | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1559010 = 111100111001102
the Final answer: 3CE616 = 111100111001102