answer:To determine which grid sizes ( m times n ) (with ( m, n geq 3 )) have a valid colouring, we need to analyze the given conditions and derive the necessary constraints. 1. **Condition Analysis**: - All squares touching the border of the grid are coloured black. - No four squares forming a ( 2 times 2 ) square are coloured in the same colour. - No four squares forming a ( 2 times 2 ) square are coloured in such a way that only diagonally touching squares have the same colour. 2. **Odd Dimensions**: - If at least one of ( m ) or ( n ) is odd, we can construct a valid colouring by alternating colours in a checkerboard pattern. This ensures that no ( 2 times 2 ) square will have the same colour in all four squares or only diagonally touching squares with the same colour. - For example, if ( m ) is odd and ( n ) is even, we can start with black on the border and alternate colours. This pattern will satisfy all conditions. 3. **Even Dimensions**: - If both ( m ) and ( n ) are even, we need to show that a valid colouring is impossible. - Consider the grid as a graph where each square is a vertex, and an edge exists between vertices if they are adjacent and have different colours. - The conditions imply that every interior vertex (square) must have exactly two edges (segments) joined to it, forming a bipartite 2-regular graph. - A bipartite 2-regular graph with an odd number of vertices is impossible because it would require an even number of vertices to maintain the bipartite property and the 2-regularity. 4. **Graph Theory Argument**: - For an ( (m-1) times (n-1) ) grid of interior vertices, if both ( m ) and ( n ) are even, the number of interior vertices is ((m-1)(n-1)), which is odd. - A bipartite 2-regular graph with an odd number of vertices cannot exist because it would require an even number of vertices to maintain the bipartite property and the 2-regularity. Therefore, the grid sizes ( m times n ) that have a valid colouring are those where at least one of ( m ) or ( n ) is odd. The final answer is ( boxed{text{At least one of } m text{ or } n text{ must be odd}} )

answer:# Step-by-Step Solution Part (1): Finding the value of p The function f(x) = |x-1| + |x+2| can be expressed in piecewise form depending on the value of x: 1. For x leq -2, we have: [ f(x) = -(x-1) - (x+2) = -2x - 1 ] 2. For -2 < x < 1, we have: [ f(x) = -(x-1) + (x+2) = 3 ] 3. For x geq 1, we have: [ f(x) = (x-1) + (x+2) = 2x + 1 ] Evaluating f(x) in each interval: - When x leq -2, f(x) = -2x - 1 geq 3. - When -2 < x < 1, f(x) = 3. - When x geq 1, f(x) = 2x + 1 geq 3. Thus, the minimum value of f(x) is p = 3. Final Answer for Part (1): [ boxed{p = 3} ] Part (2): Proving the Inequality |a+2b+3c| leqslant 6 Given a^{2}+2b^{2}+3c^{2} = 2p and from part (1), we know p=3, so: [ a^{2}+2b^{2}+3c^{2} = 6 ] Applying the Cauchy-Schwarz inequality: [ (a^{2}+2b^{2}+3c^{2})(1+2+3) geq (a+2b+3c)^{2} ] Simplifying the inequality: [ 6 times 6 geq (a+2b+3c)^{2} ] This implies: [ (a+2b+3c)^{2} leq 36 ] Taking the square root of both sides: [ |a+2b+3c| leqslant 6 ] Final Answer for Part (2): [ boxed{|a+2b+3c| leqslant 6} ] This completes the step-by-step solution for both parts of the problem.

answer:The slope k of the line x + y + 1 = 0 can be obtained by rewriting the equation in the slope-intercept form y = mx + b, where m represents the slope. Given the equation, we can rearrange it to y = -x - 1. Hence, the slope k = -1. The inclination angle alpha of a line is related to its slope through the formula tan(alpha) = k. Since our slope is k = -1, we have tan(alpha) = -1. The angle whose tangent is -1 is 135^circ (in the second quadrant where sine is positive and cosine is negative). Therefore, the inclination angle alpha of the line x + y + 1 = 0 is boxed{135^circ}.

answer:To factor 15x^2 + bx + 15 into the form (mx + n)(px + q), where m, n, p, q are integers, we equate: [ 15x^2 + bx + 15 = (mx + n)(px + q). ] Expanding, we get: [ mp x^2 + (mq + np) x + nq = 15x^2 + bx + 15. ] Comparing coefficients results in: - mp = 15 - mq + np = b - nq = 15 We explore the factorizations of 15: - (1, 15), (3, 5), (5, 3), (15, 1), and their negative counterparts. Considering equation nq = 15, permissible pairs (n, q) are same as (m, p). For example: - Using (m, p) = (3, 5) and (n, q) = (5, 3) gives: [ mq + np = 3 times 3 + 5 times 5 = 9 + 25 = 34. ] Here, b = 34 (an even number). Checking parity, since m, n, p, q must be integers and the product 15 is odd, both m, p and n, q are either both odd or both even. Hence mq + np must be even because odd times odd results in odd and odd + odd = even. Thus, any combination chosen from (1, 15), (3, 5), (5, 3), (15, 1) results in: - b always even and specific values of b enable the factorization, so b is some even number. Conclusion with boxed answer: [ textbf{(D) text{some even number}} ] The final answer is boxed{textbf{(D)} text{some even number}}