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:
3F6A16 = 3 F 6 A = 3(=0011) F(=1111) 6(=0110) A(=1010) = 111111011010102
answer: 3F6A16 = 111111011010102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+15∙162+6∙161+10∙160 = 3∙4096+15∙256+6∙16+10∙1 = 12288+3840+96+10 = 1623410
got It: 3F6A16 =1623410
Translate the number 1623410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
16234 | 2 | | | | | | | | | | | | | |
-16234 | 8117 | 2 | | | | | | | | | | | | |
0 | -8116 | 4058 | 2 | | | | | | | | | | | |
| 1 | -4058 | 2029 | 2 | | | | | | | | | | |
| | 0 | -2028 | 1014 | 2 | | | | | | | | | |
| | | 1 | -1014 | 507 | 2 | | | | | | | | |
| | | | 0 | -506 | 253 | 2 | | | | | | | |
| | | | | 1 | -252 | 126 | 2 | | | | | | |
| | | | | | 1 | -126 | 63 | 2 | | | | | |
| | | | | | | 0 | -62 | 31 | 2 | | | | |
| | | | | | | | 1 | -30 | 15 | 2 | | | |
| | | | | | | | | 1 | -14 | 7 | 2 | | |
| | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
1623410 = 111111011010102
answer: 3F6A16 = 111111011010102