answer:1. Consider the initial pile of 660 rocks. Let's assume the rocks are divided into the maximum possible number of piles. Let the smallest pile contain (m) rocks. By the problem's condition, all piles contain fewer than (2m) rocks. 2. We can look at this process in reverse: Let's combine pairs of piles to form new piles. Notice that if the current minimum number of rocks in a pile is (m), then the resulting new pile will have at most (2m) rocks. Also, once we have obtained a pile with (2m) rocks, there cannot be any remaining piles with exactly (m) rocks. 3. This means that after each combination operation, the minimum number of rocks in any pile increases. Specifically, after (k) operations, the minimum number increases by at least (k). 4. Therefore, in the original set, there were at most (2k) numbers within the interval ([m, m+k)). This implies that if we order the original set of numbers in ascending order: - The first number will be at most (m), - The third number will be at most (m+1), - The fifth number will be at most (m+2), - and so on. 5. Consequently, if (m > 15), then the first 30 numbers in the set must sum to at least: [ 15 cdot (16 + 30) = 15 cdot 46 = 690 > 660 ] However, this is a contradiction since the total number of rocks is 660. 6. On the other hand, if (m < 15), then the interval ([m, 2m)) contains less than 15 numbers. Therefore, there are less than 30 numbers in the set that are within this interval (since outside this interval there are no other numbers by the condition given). # Conclusion: Based on the above logic and contradiction, it is impossible to divide a pile of 660 rocks into 31 piles such that the size of any two piles are within twice of each other's size. (blacksquare)

answer:To find the divisor, we can use the formula for division: Dividend = (Divisor × Quotient) + Remainder We are given: Dividend = 181 Quotient = 9 Remainder = 1 Plugging these values into the formula, we get: 181 = (Divisor × 9) + 1 Now, we can solve for the Divisor: 181 - 1 = Divisor × 9 180 = Divisor × 9 Divide both sides by 9 to find the Divisor: Divisor = 180 / 9 Divisor = 20 So, the divisor is boxed{20} .

answer:1. First, consider that choosing two numbers x and y from the interval [-1, 1] forms a square region centered at the origin with edges parallel to the coordinate axes. 2. The condition x^2 + y^2 < frac{1}{4} describes a circular region centered at the origin with radius frac{1}{2}. 3. To find the probability, we need to divide the area of this circle by the area of the square. Area of the circle: pi r^2 = pi (frac{1}{2})^2 = frac{pi}{4}. Area of the square: side length 2 times 2 = 4. Probability = frac{text{Area of the circle}}{text{Area of the square}} = frac{frac{pi}{4}}{4} = boxed{frac{pi}{16}}.

answer:Part (1) We are given the quadratic function ( y = x^2 + 2mx - n^2 ). 1. **Identify Intersection Points**: - The parabola intersects the coordinate axes at three distinct points: let these points be ( A(x_1, 0) ), ( B(x_2, 0) ), and ( C(0, -n^2) ). 2. **Calculate the Product of Roots**: - By Vieta's formulas, for the quadratic ( x^2 + 2mx - n^2 = 0 ), the product of roots ( x_1 ) and ( x_2 ) is given by: [ x_1 x_2 = -n^2. ] - Clearly, ( n neq 0 ), otherwise the parabola would not intersect the axes in three distinct points. 3. **Position of Roots**: - Since ( x_1 x_2 = -n^2 < 0 ), it implies that ( x_1 ) and ( x_2 ) must have opposite signs (one is positive and the other is negative). 4. **Circle Through Points**: - Consider a point ( P_0(0, 1) ). - We examine the relationship ( |OA| cdot |OB| = |OP_0| cdot |OC| ), where ( O ) is the origin. - Given that ( x_1 ) and ( x_2 ) are roots: [ |OA| = |x_1|, quad |OB| = |x_2|, quad |OC| = n^2, quad |OP_0| = 1. ] - Therefore: [ |x_1| cdot |x_2| = n^2 cdot 1. ] - We conclude that points ( A, B, C, P_0 ) all lie on the same circle, with ( P_0(0,1) ) being a common point through which all these parabolas pass. Thus, all such parabolas must pass through the fixed point ( P_0(0,1) ). Conclusion for Part (1): [ boxed{(0, 1)} ] Part (2) 5. **Using the Given Point (1,1)**: - Since the parabola passes through ( (1, 1) ): [ 1 = 1^2 + 2m cdot 1 - n^2 ] Simplifying: [ 1 = 1 + 2m - n^2 ] [ 0 = 2m - n^2. ] Hence: [ m = frac{n^2}{2}. ] 6. **Comparison of (m) and (n+4)**: We compare ( m ) and ( n + 4 ): [ m - (n + 4) = frac{n^2}{2} - (n + 4) ] [ = frac{n^2}{2} - n - 4 ] [ = frac{1}{2} (n^2 - 2n - 8) ] [ = frac{1}{2} (n - 4)(n + 2). ] 7. **Determining (P)**: [ text{Therefore, } P = begin{cases} frac{n^2}{2} & text{if } n leq -2 text{ or } n geq 4 n + 4 & text{if } -2 < n < 4 end{cases} ] To find the minimum value of (P): - Analyze ( n = -2 ): [ P = n + 4 = -2 + 4 = 2. ] Conclusion for Part (2): [ boxed{2} ]