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+7∙162+2∙161+5∙160 = 1∙4096+7∙256+2∙16+5∙1 = 4096+1792+32+5 = 592510
got It: 172516 =592510
Translate the number 592510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
5925 | 8 | | | | |
-5920 | 740 | 8 | | | |
5 | -736 | 92 | 8 | | |
| 4 | -88 | 11 | 8 | |
| | 4 | -8 | 1 | |
| | | 3 | | |
|
the result of the conversion was:
592510 = 134458
answer: 172516 = 134458
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
172516 = 1 7 2 5 = 1(=0001) 7(=0111) 2(=0010) 5(=0101) = 10111001001012
answer: 172516 = 10111001001012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0010111001001012 = 001 011 100 100 101 = 001(=1) 011(=3) 100(=4) 100(=4) 101(=5) = 134458
answer: 172516 = 134458