This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
5∙163+10∙162+2∙161+3∙160 = 5∙4096+10∙256+2∙16+3∙1 = 20480+2560+32+3 = 2307510
got It: 5A2316 =2307510
Translate the number 2307510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
23075 | 8 | | | | |
-23072 | 2884 | 8 | | | |
3 | -2880 | 360 | 8 | | |
| 4 | -360 | 45 | 8 | |
| | 0 | -40 | 5 | |
| | | 5 | | |
|
the result of the conversion was:
2307510 = 550438
the Final answer: 5A2316 = 550438
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
5A2316 = 5 A 2 3 = 5(=0101) A(=1010) 2(=0010) 3(=0011) = 1011010001000112
the Final answer: 5A2316 = 1011010001000112
let\'s make a direct translation from binary to post-binary like this:
1011010001000112 = 101 101 000 100 011 = 101(=5) 101(=5) 000(=0) 100(=4) 011(=3) = 550438
the Final answer: 5A2316 = 550438