answer:1. **Initial Setup and Definitions**: - We have a (2007 times 2007) grid of unit squares. - Arnold's move: taking four unit squares that form a (2 times 2) square. - Bernold's move: taking a single unit square. - Arnold starts first, and they alternate turns. - When Arnold can no longer make a move, Bernold takes all remaining unit squares. 2. **Admissible Points**: - Define a point (P) as **admissible** if a (2 times 2) square can be centered at (P). - Initially, there are (2006^2) admissible points, forming a (2006 times 2006) lattice grid. 3. **Winning Condition**: - Arnold wins if he can play at least (frac{1003 cdot 1004}{2} + 1) times. - We need to show that Bernold can prevent Arnold from making this many moves. 4. **Partitioning the Grid**: - Partition the (2006 times 2006) grid into (frac{2006^2 - 2^2}{8}) subgrids of size (2 times 4) and one (2 times 2) subgrid. - This partitioning ensures that each (2 times 4) subgrid can be divided into two (2 times 2) squares. 5. **Bernold's Strategy**: - Whenever Bernold plays in a subgrid, he makes at least half of the subgrid (if it's a (2 times 4) grid) no longer admissible. - Bernold places his (1 times 1) square in the unit square bounded by the remaining (2 times 2) subgrid that is still admissible to make it un-admissible, as long as this space is vacant. - If the entire subgrid is already un-admissible, Bernold plays wherever is empty. 6. **Restricting Arnold's Moves**: - By following this strategy, Bernold ensures that Arnold can play on at most one point in each of the subgrids. - Therefore, Arnold is restricted to at most (frac{2006^2 - 2^2}{8} + 1) moves. 7. **Comparison with Winning Condition**: - We need to show that (frac{2006^2 - 2^2}{8} + 1 < frac{1003 cdot 1004}{2} + 1). - Simplifying the expressions: [ frac{2006^2 - 2^2}{8} + 1 = frac{2006^2 - 4}{8} + 1 = frac{2006^2}{8} - frac{4}{8} + 1 = frac{2006^2}{8} - 0.5 + 1 = frac{2006^2}{8} + 0.5 ] [ frac{1003 cdot 1004}{2} + 1 = frac{1003 cdot 1004}{2} + 1 ] - Since (frac{2006^2}{8} + 0.5 < frac{1003 cdot 1004}{2} + 1), Bernold can indeed prevent Arnold from making the required number of moves to win. (blacksquare)

answer:To find the length of the side of the square base, we first need to determine the volume of the marble rectangular prism. We can do this by using the weight of the prism and its density. The formula to find the volume (V) using the weight (W) and density (D) is: V = W / D We know the weight (W) is 86,400 kg and the density (D) is 2700 kg/m³, so we can plug these values into the formula: V = 86,400 kg / 2700 kg/m³ V = 32 m³ Now that we have the volume of the rectangular prism, we can use the fact that it's a rectangular prism with a square base to find the length of the side of the square base. The volume (V) of a rectangular prism is given by the formula: V = base area × height Since the base is a square, the base area is the side length squared (s²). The height (h) is given as 8 meters. So we can write: V = s² × h We know the volume (V) is 32 m³ and the height (h) is 8 m, so we can solve for the side length (s): 32 m³ = s² × 8 m Now, divide both sides by 8 m to solve for s²: 32 m³ / 8 m = s² 4 m² = s² To find the side length (s), we take the square root of both sides: s = √(4 m²) s = 2 m So, the length of the side of the square base is boxed{2} meters.

answer:The ratio of the number of individuals in the three parts is 112:16:32 = 7:1:2. Let the number of individuals drawn from each part be 7x, x, and 2x, respectively. By solving 7x + x + 2x = 20, we get x = 2. Therefore, the number of business personnel, management personnel, and logistics personnel to be drawn are 14, 2, and 4, respectively. For each part, use simple random sampling to draw individuals. If the 160 staff members are numbered sequentially as business personnel, management personnel, and logistics personnel from 1 to 160, then the individual numbers for the first part among the 112 business personnel range from 1 to 8. Randomly select a number, for example, number 4. Starting from number 4, use systematic sampling to select one number every 8 numbers. This way, the 14 numbers of the 112 business personnel drawn are as follows: 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108. Similarly, the numbers for the drawn management personnel and logistics personnel are 116, 124 and 132, 140, 148, 156, respectively. By combining the individuals drawn from each layer, a sample of 20 is obtained. boxed{text{The numbers of individuals to be drawn from business personnel, management personnel, and logistics personnel are 14, 2, and 4, respectively.}}

answer:- **Step 1: Identify series characteristics** - First term, ( a = 1 ) - Common ratio, ( r = frac{1}{3} ) - **Step 2: Apply infinite geometric series sum formula** - For |r| < 1, the sum ( S ) of the infinite geometric series is given by: [ S = frac{a}{1 - r} ] - Plugging in the values: [ S = frac{1}{1 - frac{1}{3}} = frac{1}{frac{2}{3}} = frac{3}{2} ] - So the sum of the series is ( boxed{frac{3}{2}} ). Conclusion: - The calculation above correctly applies the formula for the sum of an infinite geometric series, considering the modified common ratio ( frac{1}{3} ). The steps are logically consistent, and the expression for the sum as a common fraction is correctly simplified.