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:
2∙83+3∙82+1∙81+2∙80 = 2∙512+3∙64+1∙8+2∙1 = 1024+192+8+2 = 122610
got It: 23128 =122610
Translate the number 122610 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
1226 | 16 | | |
-1216 | 76 | 16 | |
A | -64 | 4 | |
| C | | |
|
the result of the conversion was:
122610 = 4CA16
the Final answer: 23128 = 4CA16
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
23128 = 2 3 1 2 = 2(=010) 3(=011) 1(=001) 2(=010) = 0100110010102
the Final answer: 23128 = 100110010102
Fill in the number with missing zeros on the left
let\'s do a direct translation from binary to hexadecimal like this:
0100110010102 = 0100 1100 1010 = 0100(=4) 1100(=C) 1010(=A) = 4CA16
the Final answer: 0100110010108 = 4CA16