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:
8∙163+3∙162+10∙161+3∙160 = 8∙4096+3∙256+10∙16+3∙1 = 32768+768+160+3 = 3369910
got It: 83A316 =3369910
Translate the number 3369910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
33699 | 8 | | | | | |
-33696 | 4212 | 8 | | | | |
3 | -4208 | 526 | 8 | | | |
| 4 | -520 | 65 | 8 | | |
| | 6 | -64 | 8 | 8 | |
| | | 1 | -8 | 1 | |
| | | | 0 | | |
|
the result of the conversion was:
3369910 = 1016438
answer: 83A316 = 1016438
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
83A316 = 8 3 A 3 = 8(=1000) 3(=0011) A(=1010) 3(=0011) = 10000011101000112
answer: 83A316 = 10000011101000112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010000011101000112 = 001 000 001 110 100 011 = 001(=1) 000(=0) 001(=1) 110(=6) 100(=4) 011(=3) = 1016438
answer: 83A316 = 1016438