6 Chapter 1
1.3 Parity
43 Example Two diametrically opposite corners of a chess board are deleted. Shew that it is impossible to tile the remaining
62 squares with 31 dominoes. Solution: Each domino covers one red square and one black squares. But diametrically opposite corners are of the same colour,
hence this tiling is impossible.
44 Example All the dominoes in a set are laid out in a chain according to the rules of the game. If one end of the chain is a 6,
what is at the other end? Solution: At the other end there must be a 6 also. Each number of spots must occur in a pair, so that we may put them end to
end. Since there are eight 6’s, this last 6 pairs off with the one at the beginning of the chain.
45 Example The numbers 1
, 2, . . . , 10 are written in a row. Shew that no matter what choice of sign ± is put in between them, the sum will never be 0.
Solution: The sum 1 + 2 + ··· + 10 = 55, an odd integer. Since parity is not affected by the choice of sign, for any choice of
sign ±1 ± 2 ± ···± 10 will never be even, in particular it will never be 0.
46 Definition A lattice point m
, n on the plane is one having integer coordinates.
47 Definition The midpoint of the line joining x
, y to x
1
, y
1
is the point x + x
1
2 ,
y + y
1
2 .
48 Example Five lattice points are chosen at random. Prove that one can always find two so that the midpoint of the line
joining them is also a lattice point. Solution: There are four parity patterns: even, even, even, odd, odd, odd, odd, even. By the Pigeonhole Principle among
five lattice points there must be two having the same parity pattern. Choose these two. It is clear that their midpoint is an integer.
Practice 7
For the next few examples we will need to know the names of the following tetrominoes.
Figure 1.1: L-tetromino Figure 1.2: T-tetromino
Figure 1.3: Straight-tetromino Figure 1.4: Skew-tetromino
Figure 1.5: Square-tetromino
49 Example A single copy of each of the tetrominoes shewn above is taken. Shew that no matter how these are arranged, it is
impossible to construct a rectangle. Solution: If such a rectangle were possible, it would have 20 squares. Colour the rectangle like a chessboard. Then there are 10
red squares and 10 black squares. The T-tetromino always covers an odd number of red squares. The other tetrominoes always cover an even number of red squares. This means that the number of red squares covered is odd, a contradiction.
50 Example Shew that an 8
× 8 chessboard cannot be tiles with 15 straight tetrominoes and one L-tetromino. Solution: Colour rows 1
, 3, 5, 7 black and colour rows 2, 4, 6, and 8 red. A straight tetromino will always cover an even number of red boxes and the L-tetromino will always cover an odd number of red squares. If the tiling were possible, then we would be
covering an odd number of red squares, a contradiction.
Practice
51 Problem Twenty-five boys and girls are seated at a round table. Shew that both
neighbours of at least one student are girls.
52 Problem A closed path is made of 2001 line segments. Prove that there is no line,
not passing through a vertex of the path, intersecting each of the segments of the path.
53 Problem The numbers 1
, 2, . . . ,2001 are written on a blackboard. One starts erasing any two of them and replacing the deleted ones with their difference. Will a situation
arise where all the numbers on the blackboard be 0?
54 Problem Shew that a 10
× 10 chessboard cannot be tiled with 25 straight tetromi- noes.
55 Problem Shew that an 8
× 8 chess board cannot be tiled with 15 T-tetrominoes and one square tetromino.
Chapter
2
Algebra
2.1 Identities with Squares
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