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:
10∙163+1∙162+4∙161+12∙160 = 10∙4096+1∙256+4∙16+12∙1 = 40960+256+64+12 = 4129210
got It: A14C16 =4129210
Translate the number 4129210 в octal like this:
the Integer part of the number is divided by the base of the new number system:
41292 | 8 | | | | | |
-41288 | 5161 | 8 | | | | |
4 | -5160 | 645 | 8 | | | |
| 1 | -640 | 80 | 8 | | |
| | 5 | -80 | 10 | 8 | |
| | | 0 | -8 | 1 | |
| | | | 2 | | |
|
the result of the conversion was:
4129210 = 1205148
answer: A14C16 = 1205148
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
A14C16 = A 1 4 C = A(=1010) 1(=0001) 4(=0100) C(=1100) = 10100001010011002
answer: A14C16 = 10100001010011002
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010100001010011002 = 001 010 000 101 001 100 = 001(=1) 010(=2) 000(=0) 101(=5) 001(=1) 100(=4) = 1205148
answer: A14C16 = 1205148