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:
15∙165+1∙164+14∙163+2∙162+12∙161+3∙160 = 15∙1048576+1∙65536+14∙4096+2∙256+12∙16+3∙1 = 15728640+65536+57344+512+192+3 = 1585222710
got It: F1E2C316 =1585222710
Translate the number 1585222710 в octal like this:
the Integer part of the number is divided by the base of the new number system:
15852227 | 8 | | | | | | | |
-15852224 | 1981528 | 8 | | | | | | |
3 | -1981528 | 247691 | 8 | | | | | |
| 0 | -247688 | 30961 | 8 | | | | |
| | 3 | -30960 | 3870 | 8 | | | |
| | | 1 | -3864 | 483 | 8 | | |
| | | | 6 | -480 | 60 | 8 | |
| | | | | 3 | -56 | 7 | |
| | | | | | 4 | | |
|
the result of the conversion was:
1585222710 = 743613038
answer: F1E2C316 = 743613038
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
F1E2C316 = F 1 E 2 C 3 = F(=1111) 1(=0001) E(=1110) 2(=0010) C(=1100) 3(=0011) = 1111000111100010110000112
answer: F1E2C316 = 1111000111100010110000112
let\'s make a direct translation from binary to post-binary like this:
1111000111100010110000112 = 111 100 011 110 001 011 000 011 = 111(=7) 100(=4) 011(=3) 110(=6) 001(=1) 011(=3) 000(=0) 011(=3) = 743613038
answer: F1E2C316 = 743613038