answer:To solve the problem, we need to ensure the sum of the numbers in any four-cell square within the (11 times 11) grid is odd. The only way this can be achieved is if one or three of the cells in every four-cell square contains a (1). 1. **Dividing the Grid into Non-Overlapping Squares:** - We divide the (11 times 11) grid into (2 times 2) subgrids. Since each (2 times 2) square must contain at least one (1), we count these non-overlapping (2 times 2) squares. - The number of horizontal (2 times 2) squares is (leftlfloor frac{11}{2} rightrfloor = 5) - The number of vertical (2 times 2) squares is (leftlfloor frac{11}{2} rightrfloor = 5) - Therefore, in total, there are (5 times 5 = 25) non-overlapping (2 times 2) subgrids in the (11 times 11) grid. 2. **Ensuring Each Square Contains Exactly One (1):** - Each of these (2 times 2) squares needs to have at least one (1) to ensure any such four-cell square's sum is odd. - The smallest number of (1)s necessary is therefore one (1) per each (2 times 2) square, leading to at least (25) (1)s for the entire grid. 3. **Determining a Suitable Arrangement:** - We place a (1) in every cell with even coordinates within the grid. Even coordinates ensure each (2 times 2) square gets one (1) as follows: [ text{Positioning in even coordinates: } (2k, 2l), text{ where } k text{ and } l text{ are integers from } {0, 1, ldots, 5} ] - These coordinates fill up one cell in each (2 times 2) block of the (11 times 11) grid. 4. **Conclusion:** - Ensuring at least one (1) in each (2 times 2) square guarantees that all possible configurations of four-cell squares sum to an odd number. - Therefore, the minimum number of (1)s is (boxed{25}). Hence, the minimum number of (1)s required for this configuration is: (boxed{25})

answer:First, we factor out (a-b): [ a^4 (b^2 - c^2) + b^4 (c^2 - a^2) + c^4 (a^2 - b^2) = a^4 b^2 - a^4 c^2 - a^2 b^4 + b^4 c^2 + c^4 a^2 - c^4 b^2. ] Then, [ = a^2 a^2 b^2 - a^4 c^2 - b^2 b^2 a^2 + b^4 c^2 + c^2 c^2 a^2 - c^4 b^2 ] [ = a^2 b^2 (a^2 - b^2) + b^4 c^2 - a^4 c^2 + c^4 a^2 - c^4 b^2. ] [ = (a - b)(a + b)a^2b^2 + b^4 c^2 - a^4 c^2 + c^4 a^2 - c^4 b^2. ] Factor out (b-c): [ = (a - b)(a+b)a^2b^2 + b^2(b^2 - c^2) + c^2(c^2 - a^2). ] [ = (a - b)(a + b)a^2b^2 + (b-c)(b+c)b^2c^2 + c^2(-a+c)(-a-c). ] Finally, factor out (c-a): [ = (a - b)(a + b)a^2b^2 + (b-c)(b+c)b^2c^2 + (c-a)(a+c)c^2a. ] Thus, [ q(a,b,c) = boxed{(a+b)a^2b^2 + (b+c)b^2c^2 + (a+c)c^2a}. ]

answer:According to the property of a quadrilateral inscribed in a circle, we have (a+2)(a-1) + (1-a)(2a+3) = 0. Simplifying this equation, we get a^2 = 1. Therefore, a = pm 1. So, the final answer is boxed{a = pm 1}.

answer:**Analysis** This problem involves the graph and properties of a quadratic function, requiring both transformational thinking and simplification skills. From the given conditions, we can deduce that a geqslant 1. The condition for the quadratic equation to have real roots is that its discriminant is non-negative. By eliminating c and using the absolute value inequality solution method, combined with the given conditions, we can find the range of a and hence its minimum value. **Solution** Given a + b + c = 3, a geqslant b geqslant c, Hence, 3a geqslant 3, i.e., a geqslant 1, Since ax^{2}+bx+c=0 has real roots, Hence, Delta = b^{2}-4ac geqslant 0, Substituting c = 3 - a - b into b^{2}-4ac geqslant 0 yields, (b + 2a)^{2} geqslant 12a, This implies b geqslant 2sqrt{3a} - 2a or b leqslant -2sqrt{3a} - 2a, So we must have a geqslant 2sqrt{3a} - 2a or 3 - a - c leqslant -2sqrt{3a} - 2a, This simplifies to 3 + a + 2sqrt{3a} leqslant c, Given a geqslant c and a geqslant 1, Hence, 3 + a + 2sqrt{3a} leqslant c is not valid, so we discard it; Thus, 3a geqslant 2sqrt{3a}, which gives a geqslant frac{4}{3}, Therefore, the minimum value of a is boxed{frac{4}{3}}.