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:
3∙162+14∙161+7∙160 = 3∙256+14∙16+7∙1 = 768+224+7 = 99910
got It: 3E716 =99910
Translate the number 99910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
999 | 8 | | | |
-992 | 124 | 8 | | |
7 | -120 | 15 | 8 | |
| 4 | -8 | 1 | |
| | 7 | | |
|
the result of the conversion was:
99910 = 17478
answer: 3E716 = 17478
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
3E716 = 3 E 7 = 3(=0011) E(=1110) 7(=0111) = 11111001112
answer: 3E716 = 11111001112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011111001112 = 001 111 100 111 = 001(=1) 111(=7) 100(=4) 111(=7) = 17478
answer: 3E716 = 17478