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:
4∙162+14∙161+7∙160 = 4∙256+14∙16+7∙1 = 1024+224+7 = 125510
got It: 4E716 =125510
Translate the number 125510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
1255 | 8 | | | |
-1248 | 156 | 8 | | |
7 | -152 | 19 | 8 | |
| 4 | -16 | 2 | |
| | 3 | | |
|
the result of the conversion was:
125510 = 23478
answer: 4E716 = 23478
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
4E716 = 4 E 7 = 4(=0100) E(=1110) 7(=0111) = 100111001112
answer: 4E716 = 100111001112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0100111001112 = 010 011 100 111 = 010(=2) 011(=3) 100(=4) 111(=7) = 23478
answer: 4E716 = 23478