This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s translate to decimal like this:
4∙82+5∙81+3∙80 = 4∙64+5∙8+3∙1 = 256+40+3 = 29910
got It: 4538 =29910
Translate the number 29910 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
299 | 16 | | |
-288 | 18 | 16 | |
B | -16 | 1 | |
| 2 | | |
|
the result of the conversion was:
29910 = 12B16
the Final answer: 4538 = 12B16
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
4538 = 4 5 3 = 4(=100) 5(=101) 3(=011) = 1001010112
the Final answer: 4538 = 1001010112
Fill in the number with missing zeros on the left
let\'s do a direct translation from binary to hexadecimal like this:
0001001010112 = 0001 0010 1011 = 0001(=1) 0010(=2) 1011(=B) = 12B16
the Final answer: 0001001010118 = 12B16