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:
6F2A16 = 6 F 2 A = 6(=0110) F(=1111) 2(=0010) A(=1010) = 1101111001010102
the Final answer: 6F2A16 = 1101111001010102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
6∙163+15∙162+2∙161+10∙160 = 6∙4096+15∙256+2∙16+10∙1 = 24576+3840+32+10 = 2845810
got It: 6F2A16 =2845810
Translate the number 2845810 в binary like this:
the Integer part of the number is divided by the base of the new number system:
28458 | 2 | | | | | | | | | | | | | | |
-28458 | 14229 | 2 | | | | | | | | | | | | | |
0 | -14228 | 7114 | 2 | | | | | | | | | | | | |
| 1 | -7114 | 3557 | 2 | | | | | | | | | | | |
| | 0 | -3556 | 1778 | 2 | | | | | | | | | | |
| | | 1 | -1778 | 889 | 2 | | | | | | | | | |
| | | | 0 | -888 | 444 | 2 | | | | | | | | |
| | | | | 1 | -444 | 222 | 2 | | | | | | | |
| | | | | | 0 | -222 | 111 | 2 | | | | | | |
| | | | | | | 0 | -110 | 55 | 2 | | | | | |
| | | | | | | | 1 | -54 | 27 | 2 | | | | |
| | | | | | | | | 1 | -26 | 13 | 2 | | | |
| | | | | | | | | | 1 | -12 | 6 | 2 | | |
| | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
2845810 = 1101111001010102
the Final answer: 6F2A16 = 1101111001010102