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:
1∙163+2∙162+1∙161+4∙160 = 1∙4096+2∙256+1∙16+4∙1 = 4096+512+16+4 = 462810
got It: 121416 =462810
Translate the number 462810 в octal like this:
the Integer part of the number is divided by the base of the new number system:
4628 | 8 | | | | |
-4624 | 578 | 8 | | | |
4 | -576 | 72 | 8 | | |
| 2 | -72 | 9 | 8 | |
| | 0 | -8 | 1 | |
| | | 1 | | |
|
the result of the conversion was:
462810 = 110248
answer: 121416 = 110248
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
121416 = 1 2 1 4 = 1(=0001) 2(=0010) 1(=0001) 4(=0100) = 10010000101002
answer: 121416 = 10010000101002
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010010000101002 = 001 001 000 010 100 = 001(=1) 001(=1) 000(=0) 010(=2) 100(=4) = 110248
answer: 121416 = 110248