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:
9E1A16 = 9 E 1 A = 9(=1001) E(=1110) 1(=0001) A(=1010) = 10011110000110102
answer: 9E1A16 = 10011110000110102
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
9∙163+14∙162+1∙161+10∙160 = 9∙4096+14∙256+1∙16+10∙1 = 36864+3584+16+10 = 4047410
got It: 9E1A16 =4047410
Translate the number 4047410 в binary like this:
the Integer part of the number is divided by the base of the new number system:
40474 | 2 | | | | | | | | | | | | | | | |
-40474 | 20237 | 2 | | | | | | | | | | | | | | |
0 | -20236 | 10118 | 2 | | | | | | | | | | | | | |
| 1 | -10118 | 5059 | 2 | | | | | | | | | | | | |
| | 0 | -5058 | 2529 | 2 | | | | | | | | | | | |
| | | 1 | -2528 | 1264 | 2 | | | | | | | | | | |
| | | | 1 | -1264 | 632 | 2 | | | | | | | | | |
| | | | | 0 | -632 | 316 | 2 | | | | | | | | |
| | | | | | 0 | -316 | 158 | 2 | | | | | | | |
| | | | | | | 0 | -158 | 79 | 2 | | | | | | |
| | | | | | | | 0 | -78 | 39 | 2 | | | | | |
| | | | | | | | | 1 | -38 | 19 | 2 | | | | |
| | | | | | | | | | 1 | -18 | 9 | 2 | | | |
| | | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
4047410 = 10011110000110102
answer: 9E1A16 = 10011110000110102