answer:Let C have coordinates (p, q). Then by the Shoelace Formula, the area of triangle ABC is frac{1}{2} lvert 10p - 24q rvert. Since p and q are integers, lvert 10p - 24q rvert is a positive integer. By Bezout's Lemma, the greatest common divisor (gcd) of 10 and 24 is 2, so the smallest lvert 10p - 24q rvert can be is 2. Thus, the minimum area is frac{1}{2} times 2 = boxed{1}.

answer:# Problem: 为适应市场竞争，某品牌方便面实行 “加量不加价” 的销售策略。具体实施办法为: 给每袋方便面都增加 (25 %) 的重量，但每袋方便面的售价保持不变；那么这相当于每袋方便面降价 (quad %) 销售。 To determine the equivalent percentage drop in price due to the increase in weight: 1. Let the original weight of the instant noodle be ( W ), 2. Let the original price be ( P ). Given that the weight increases by ( 25 % ), the new weight is: [ W_{text{new}} = W + 0.25W = 1.25W ] The price remains ( P ). Thus, the new price per unit weight is: [ text{Price per unit weight (new)} = frac{P}{1.25W} = frac{P}{1.25} cdot frac{1}{W} = frac{P}{W} cdot frac{1}{1.25} ] Simplifying (frac{1}{1.25}): [ frac{1}{1.25} = frac{1}{frac{5}{4}} = frac{4}{5} = 0.8 ] Hence, the new price per unit weight is: [ text{Price per unit weight (new)} = frac{P}{W} cdot 0.8 ] Now, comparing the new price per unit weight to the original price per unit weight: [ text{Percentage decrease} = 1 - 0.8 = 0.2 = 20 % ] This shows that the price per unit weight is effectively reduced by 20%. Conclusion: [ boxed{20} ]

answer:Let's approach this problem step-by-step: 1. **Problem Setup**: We have an infinite chessboard with tokens placed every three squares both horizontally and vertically. We need to investigate whether a knight can visit every remaining square exactly once. 2. **Consideration of Token Placement**: Assume tokens are placed on black squares (without loss of generality). 3. **Structure Analysis**: Consider a sufficiently large square of size ( (4N-3) times (4N-3) ) (we will define ( N ) more precisely later). Let's first analyze the squares: - Place tokens in the corners of this square, and - Expand it to a square of size ( (4N+1) times (4N+1) ), achieved by adding 2 rows and 2 columns to each side of the original square. The total number of squares in ( (4N-3) times (4N-3) ) is: [ (4N-3)(4N-3) ] 4. **Counting the White Squares**: The original square ( (4N-3) times (4N-3) ) will contain: [ A(N) = frac{(4N-3)(4N-3) - 1}{2} = 8N^2 - 12N + 4 ] white squares. Here, we subtract 1 to account for the extra square when counting. 5. **Knight Movement Considerations**: If the knight is to cover all non-token squares, it will move from each white square to a distinct black square in the expanded square ( (4N+1) times (4N+1) ) without tokens. 6. **Counting the Black Squares in the Expanded Square**: The total number of black squares in the expanded square ( (4N+1) times (4N+1) ): [ B(N) = frac{(4N+1)(4N+1) + 1}{2} - N^2 = 7N^2 + 4N + 1 ] Here, we add 1 for the correct parity balance and subtract ( N^2 ) as those are occupied by tokens. 7. **Analyzing the Requirement**: For the knight to visit each remaining square, ( B(N) ) must be at least ( A(N) ): [ A(N) - B(N) = (8N^2 - 12N + 4) - (7N^2 + 4N + 1) = N^2 - 16N + 3 ] 8. **Checking the Condition for Large ( N )**: Consider large ( N ): [ N^2 - 16N + 3 > 0 quad text{for} quad N > 16 ] 9. **Conclusion**: Since for ( N > 16 ), ( A(N) ) exceeds ( B(N) ), we get a contradiction showing that the knight cannot visit each remaining square exactly once given the token placement. Therefore, it is impossible for the knight to achieve the required tour. Conclusion: [ boxed{text{Не можно}} ]

answer:1. **Initial Simplification**: Simplify the expression for G: G = frac{8x^2 + 24x + 3n}{8} = x^2 + 3x + frac{3n}{8} 2. **Perfect Square Requirement**: For G to be the square of a linear expression, it should take the form (cx + d)^2. Expanding this gives: (cx + d)^2 = c^2x^2 + 2cdx + d^2 By comparing this with x^2 + 3x + frac{3n}{8}, we can equate: c^2 = 1, quad 2cd = 3, quad d^2 = frac{3n}{8} 3. **Solving for Coefficients**: From c^2 = 1, we find c = 1 or c = -1. Assuming c = 1: 2(1)d = 3 implies d = frac{3}{2} Substituting d = frac{3}{2} into d^2 = frac{3n}{8}: left(frac{3}{2}right)^2 = frac{3n}{8} implies frac{9}{4} = frac{3n}{8} implies n = frac{9}{4} times frac{8}{3} = 6 Conclusion: The value of n that satisfies the condition is 6. The final answer is boxed{(D) 6}