Odds and evens

fetofs · 2007-03-24, 00:55

You have three sequences A = a₁, a₂,...,a_n, B = b₁, b₂,...,b_n and C = c₁, c₂,...,c_n. For each 1 <= i <= n it is known that at least one of a_i, b_i and c_i is odd. Prove that there are integers r, s and t such that ra_i + sb_i + tc_i is odd for at least $\frac{4n}7$ values of i.

R. Gerbicz · 2007-03-24, 15:46

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.

2007-03-24, 00:55	#1
fetofs Aug 2005 Brazil 2×181 Posts	Odds and evens You have three sequences A = a₁, a₂,...,a_n, B = b₁, b₂,...,b_n and C = c₁, c₂,...,c_n. For each 1 <= i <= n it is known that at least one of a_i, b_i and c_i is odd. Prove that there are integers r, s and t such that ra_i + sb_i + tc_i is odd for at least $\frac{4n}7$ values of i.

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2007-03-24, 15:46	#2
R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 2·743 Posts	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<4/7n is true. Similarly: Let r=0,s=1,t=0, then p2+p3+p6+p7<4/7n Let r=0,s=1,t=1, then p1+p2+p5+p6<4/7m Let r=1,s=0,t=0, then p4+p5+p6+p7<4/7n Let r=1,s=0,t=1, then p1+p3+p4+p6<4/7n Let r=1,s=1,t=0, then p2+p3+p4+p5<4/7n Let r=1,s=1,t=1, then p1+p2+p4+p7<4/7n 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.