Aufgabe #13
#13 is not hard.
Take all possible 6-letter codes, and add to each combination 7th letter
computed as sum of first 6 letters by modulo 7 (assuming A=0, ..., G=6).
So we have 7**6 codes.
Example in the text of the problem is quite strong hint
Aufgabe #7.
The answer is 24 (not 32).
First:
AAAA, AABB, ABAB, ABBA (4 codes)
and replace in each code A<-->B (more 4 codes)
These 8 codes cover all 80 combinations where total count of A and B is greater than 2.
Now more tricky:
x(AB)y - denote 2 codes xAy and xBy
C(AB)CD
CD(AB)C
CCD(AB)
(AB)CCC
(8 codes)
and replace in each code C<-->D (more 8 codes)
Each from 8 pairs x(AB)Y cover 22 combinations
Total 80+8*22=256 combinations.
Check it out yourself.
Beautiful, is not it?
#13 is not hard.
Take all possible 6-letter codes, and add to each combination 7th letter
computed as sum of first 6 letters by modulo 7 (assuming A=0, ..., G=6).
So we have 7**6 codes.
Example in the text of the problem is quite strong hint
Aufgabe #7.
The answer is 24 (not 32).
First:
AAAA, AABB, ABAB, ABBA (4 codes)
and replace in each code A<-->B (more 4 codes)
These 8 codes cover all 80 combinations where total count of A and B is greater than 2.
Now more tricky:
x(AB)y - denote 2 codes xAy and xBy
C(AB)CD
CD(AB)C
CCD(AB)
(AB)CCC
(8 codes)
and replace in each code C<-->D (more 8 codes)
Each from 8 pairs x(AB)Y cover 22 combinations
Total 80+8*22=256 combinations.
Check it out yourself.
Beautiful, is not it?
