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:
3∙164+4∙163+5∙162+10∙161+12∙160 = 3∙65536+4∙4096+5∙256+10∙16+12∙1 = 196608+16384+1280+160+12 = 21444410
got It: 345AC16 =21444410
Translate the number 21444410 в octal like this:
the Integer part of the number is divided by the base of the new number system:
214444 | 8 | | | | | |
-214440 | 26805 | 8 | | | | |
4 | -26800 | 3350 | 8 | | | |
| 5 | -3344 | 418 | 8 | | |
| | 6 | -416 | 52 | 8 | |
| | | 2 | -48 | 6 | |
| | | | 4 | | |
|
the result of the conversion was:
21444410 = 6426548
the Final answer: 345AC16 = 6426548
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
345AC16 = 3 4 5 A C = 3(=0011) 4(=0100) 5(=0101) A(=1010) C(=1100) = 1101000101101011002
the Final answer: 345AC16 = 1101000101101011002
let\'s make a direct translation from binary to post-binary like this:
1101000101101011002 = 110 100 010 110 101 100 = 110(=6) 100(=4) 010(=2) 110(=6) 101(=5) 100(=4) = 6426548
the Final answer: 345AC16 = 6426548