Mathematical puzzles online - Links to books and blogs

1. Sample puzzles of Martin Gardner, with solutions - The Guardian
2. Martin Gardner's best math puzzles - part 1 (The Math Dude)
3. Math questions (solvable online) for class 10 from NSTE
4. International Mathematical Olympiad - Question paper archive
5. Problem Solving Strategies - Arthur Engel - archive.org
6. Entertaining Math Puzzles, by Martin Gardner
7. Math Proof Training and Practice Forum
   
   Source https://www.physicsforums.com/forums/math-proof-training-and-practice.296
   Math Proof Training and Practice Forum at PhysicsForums.com
   

2 problems - Matrix numbering and integer colouring

Here are 2 interesting problems from chapter 2 of Arthur Engel's book:

1. The positive integers are colored black and white. The sum of two differently colored
numbers is black, and their product is white. What is the product of two white
numbers? Find all such colorings.

2. Each element of a 25 × 25 matrix is either +1 or −1. Let ai be the product of all
elements of the i-th row and bj be the product of all elements of the j-th column.
Prove that a₁ + b₁ + · · · + a₂₅ + b₂₅ ≠ 0.

Tile colouring problem (#8) from Problem solving strategies

Problem: Prove that an a × b rectangle can be covered by 1 × n rectangles iff n|a or n|b.

Solution: The proof is given by means of the following 2 claims.

Claim 1: Given a rectangular area is covered by tiles of dimension 1xn placed in horizontal or vertical orientation, it is possible to remove the tiles one at a time in such a way that at any point of time, the covered region’s top border consists of a sequence of steps increasing from left to right as shown in diagram 1.

Proof: Define step blocks to be rectangular regions defined by the step-like shape of the covered area. In diagram 1, step blocks are represented by the areas shaded blue. Start from the left most step block and consider the top left tile in that block. This tile must fall entirely within the current block (call it block A) or must extend beyond the right of the current step block (it cannot beyond the bottom of the current block since the first block is also the bottommost). In the former case, removing the tile will leave behind a step shape, thereby proving the claim. In the latter case, inspect the next step block for the possibility of removing a tile. The top left tile of this next block (call it block B) cannot extend beyond the bottom of that block since the block A’s top left tile occupies that portion as per the previous statement. Hence the top left tile of block B too must either lie entirely within B or extend beyond the right of B. As before, in the former case, removing the tile will leave a step shape whereas in the latter case, the search for a tile to remove can be repeated with the block to the right of C. Note that this procedure can be repeated at most till the rightmost block is reached, in which case the top left tile cannot extend beyond the right of the block (and must therefore lie entirely within that block), thereby becoming eligible for removal while retaining the step shape.

Claim 2: When tiles are removed as per the procedure of claim 1 and neither a not b are divisible by n, then at any point of time at least one of the step regions on the top border will have a height and a width (as defined by diagram 2) that are both indivisible by n.

Proof: The proof is by induction on r, the number of tiles removed so far under the procedure of claim 1. When r is 0, the covered region and also the only step region is the original axb rectangle. The height and width, namely a and b, are both indivisible by n as per the assumption. Suppose the claim is true for values of r upto R. Now consider r = (R + 1). Suppose the tile being removed at the current iteration lies in a step region whose width is W and height Is H. If either W or H is divisible by n, then there must be some other step region with width and height indivisible by n and that region will remain after iteration (R + 1)  as well. But if say W and H are both indivisible by n. After removing the tile in this iteration, the 2 new step regions would have dimensions (a) (W, H - 1) and (W - n, H) or (b) (W, H - n) and (W - 1, H), depending on the orientation of the tile; in either case, one of the new step regions would have its width and height indivisible by n. Hence proved.

Diagram 1: Step blocks denoted by shaded blue regions

Diagram 2: Width of step regions are denoted by L1 and L2; heights are denoted by H1 and H2. The rectangle  bounded by blue and red lines is the top left tile of its step block and is eligible for removal as per the procedure of claim 1

A simple number theory problem

Divisibility puzzle: Suppose that there are n integers which have the following
property: the difference between the product of any n - 1 integers and
the remaining one is divisible by n . Prove that the sum of the square of
these n numbers is also divisible by n.

Solution:

Denote the n integers by a₁, a₂, a₃, .., a_n. By the given property, it follows that:

(a₁ – a₂ a₃…a_n) = k₁n

(a₂ – a₁ a₃ ...a_n) = k₂n

…

(a_n – a₁ a₂… a_n-1) = k_nn

where k₁, k₂, .., k_n are some integers. By multiplying equation 1 above by a₁, similarly equation 2 by a₂ and so on till the last equation above (which would be multiplied by a_n), the following set of equations are obtained:

a₁² – C = k₁na₁

a₂² – C = k₂na₂

...

a_n² – C = k_nna_n

where C = a₁ a₂ a₃ ...a_n.

Summing the above equations and then rearranging leads to:

(a₁²+ a₂² + … + a_n²) = nC  + n(k₁a₁ + k₂a₂ + … + k_na_n)  = n(C + k₁a₁ + k₂a₂ + … + k_na_n)

which proves that the sum of squares of the n integers (a₁, a₂, a₃, .., a_n) is a multiple of n.

Colouring problem #24 - 3-colour space

Problem: Every space point is colored with exactly one of the colors red, green or blue. The sets R, G, B consist of the lengths of those segments in space with both end points red, green and blue, respectively. Show that at least one of these sets contains all non-negative real numbers.

Solution:
(With reference to the diagram) Suppose the 3 colours are C1, C2 and C3. Assume that the minimum distances by which no two points with the same colour are separated (i.e. no line segment with end points of the same colour exist) be x, y and z for C1, C2 and C3 respectively. Suppose x >= y >= z. Then as shown in the diagram, any point coloured C1 would form the centre of a sphere of radius x such that the sphere can have only points coloured C2 or C3. As indicated in the diagram, a circle of diameter x√3 would then exist on the surface of the sphere such that all its points are coloured C3 and the chords of the circle form line segments of all possible lengths upto x√3, contradicting the assumption that no two points coloured C3 exist at a distance of z from each other.

Diagram for problem #24: The green circle is formed by points on surface of the sphere that are at a distance of x from the top point (on spherical surface) coloured C2, where x is the radius of the sphere centered at a point coloured C1. The diameter of the circle is xv3. If neither of the colours C1 and C2 have 2 points at a distance of x, then the circle must be coloured C3 entirely.