"Mathematical olympiad--Combinatorial Mathematics"--dyeing problem

resuming The analysis of the problems in the mathematical olympiad--combinatorial mathematics, this article will cover the problem of staining.

Question one:

Put some stones in a rectangular grid of 10 rows and 14 columns, allowing more than 1 stones to be placed in each cell, and then discovering that each row has an odd block of stones on each column. If the cells on the rectangular grid are dyed black and white, it proves that the number of stones on the black cell is a total of even blocks.

Analysis: We consider using contradiction to complete the proof. That is, the number of stones on the black cell has odd blocks.

We first assume that the rectangular square odd row ciriè, even row even column is black, and set Chickange column has K1 odd number of stone lattice, even row even column has K2 odd number of stone lattice, odd even column has K3 odd number of lattice.

There are 10 rows of odd pieces of stone in each row, with 5 odd lines, and the number of odd rows of stones is odd, with K1 + k3≡1 (mod 2) ①

Similarly, each column has an odd block of stones, a total of 14 columns, a total of 7 even numbers, but the number of stones in the even sequence is odd, with K3 + k2≡1 (mod 2) ②

Based on the hypothesis of the disprove process, K1 + k2≡1 (mod 2) ③

①+②+③, with 2 (K1+K2+K3) ≡1 (mod 2), is obviously impossible, so the hypothesis is not tenable.

Similarly, if we assume that the matrix squares odd even, even row ciriè are black, and do the same analysis, when the variables are set, the corresponding changes need to be made and the same conclusion can be obtained.

The certificate is completed.

"Mathematical olympiad--Combinatorial Mathematics"--dyeing problem