1. At least 1 statement among these 2n+1 are true.
2. At least 3 statements among these 2n+1 are false.
3. At least 5 statements among these 2n+1 are true.
...
2n. At least (4n-1) statments among these 2n+1 are false.
2n+1. At least (4n+1) statements among these 2n+1 are true.

How many statements are true? Which?

(St = Statement; V = True; F = False) (n as in the sequence)

As I said in my first mail, St from n+1 to 2n + 1 are always F (they are impossible).

Then, some of the first even St will be V (and this also assures that the first St is necessarily V).

More: some odd and even St, placed just before the St number n, will always be F. Some odd St are F because the number of VSt has to be always inferior to n, and some of this St rule against it; some even St are also F, because the number of F St can't be so close to 2n+1, that there is no room for V St (some St are F because rule against this).

All of it explain the shape of each set of 2n+1 St. The shape is the same for the different sets, with some V at the beginning of the sequence, followed by V and F alternate and, at least, a lot of F terms.  (Shape for n=13, for ex., is:

VVVVFVFVFVFFF FFFFFFFFFFFFFF (1-13 14-27)

(And for n=7) :  VVFVFVF FFFFFFFF (1-7 8-15)

I tried to obtain a general math formula, based on the process of obtention for V and F for each set, but I couldn't found an easy way. Anyway, I could noted that probably the sets of 2n+1 St, follows a 8-cycle solution, depending on the value assigned to n. As I couldn't do it with maths, I tried directly to solve some of the 2n+1 sets of St, to get inducted formulas.

This are the results I get:

If n= 8k  (=8, 16, 24, ....) there are two possible solutions(because the last assigned St admit both possibilities, V and F)

If n= 8k-1  (=7, 15, 23...)   V= (n+1)/2  F= (3n+1)/2

If n= 8k-2  (=6, 14,  22...)  V= n/2   F= (3n+2)/2

If n= 8k-3  (=5, 13,...)        (as for 8k-1)

If n= 8k-4 (= 4, 12, 20....) there is not a logical solution: assigning F to the last St convert the St automatically in V, and viceversa.

If n= 8k+1 (= 9, 17, ...)      (as for 8k-1)

If n= 8k+2  (= 10, 18,...)     V= (n+2)/2  F= 3n/2

If n= 8k+3 (=11, 19,...)       (as for 8k-1)

=========================

Examples: If n= 20 there is not a solution

If n = 37  (8*5-3)   V= (37+1)/2 = 19  F= (3*37+1)/2 = 56

I've checked a little number of sequences anyway, so I'm not all sure of all this formulas.

 

Edited on June 14, 2005, 10:25 pm

Posted by armando on 2005-06-14 16:28:08