Thread: Odds and evens
View Single Post
Old 2007-03-24, 15:46   #2
R. Gerbicz
 
R. Gerbicz's Avatar
 
"Robert Gerbicz"
Oct 2005
Hungary

3×11×43 Posts
Default


pattern number of combinations:
of (a%2,b%2,c%2):
1,1,1 p7
1,1,0 p6
1,0,1 p5
1,0,0 p4
0,1,1 p3
0,1,0 p2
0,0,1 p1

We know that: p1+p2+p3+p4+p5+p6+p7=n

Indirectly suppose that there is no good r,s,t integer values


Let r=0,s=0,t=1, is bad if and only if
p1+p3+p5+p7<4/7*n is true. Similarly:
Let r=0,s=1,t=0, then p2+p3+p6+p7<4/7*n
Let r=0,s=1,t=1, then p1+p2+p5+p6<4/7*m
Let r=1,s=0,t=0, then p4+p5+p6+p7<4/7*n
Let r=1,s=0,t=1, then p1+p3+p4+p6<4/7*n
Let r=1,s=1,t=0, then p2+p3+p4+p5<4/7*n
Let r=1,s=1,t=1, then p1+p2+p4+p7<4/7*n
Add these 7 inequalities:
4*(p1+p2+p3+p4+p5+p6+p7)<4*n So:

p1+p2+p3+p4+p5+p6+p7<n but this is a contradiction.
R. Gerbicz is offline   Reply With Quote