answer:1. **Understand the Problem**: - We have a labyrinth composed of ( n ) circles, each touching a straight line ( AB ) at point ( M ). - All circles lie on one side of the straight line. - The lengths of the circles are in geometric progression with a common ratio of 2. 2. **Circles and Geometric Progression**: - Assume the radius of the smallest circle is ( r ). - The radii of the circles then follow the progression: ( r, 2r, 4r, ldots, 2^{n-1}r ). 3. **Travel Paths of Individuals**: - Each of the two persons starts walking through the labyrinth at different times but at the same speed, though in opposite directions. - Each person follows the sequence of the circles in a given order (largest to smallest and vice versa), looping continuously. 4. **Meeting Conditions**: - To prove they will necessarily meet, we need to consider the dynamics when one person is on the largest circle. 5. **Simplifying the Argument**: - Let ( C_i ) denote the circle with radius ( 2^i r ). Here ( 0 leq i leq n-1 ). 6. **People on the Largest Circle**: - Consider the situation when one person is walking on the largest circle ( C_{n-1} ). - During the time the person traverses ( C_{n-1} ): 1. The total circumference traversed is ( 2 pi cdot 2^{n-1} r ). 7. **Comparison with Other Circles**: - The summed circumference of all smaller circles is: [ sum_{i=0}^{n-2} 2 pi cdot 2^i r = 2 pi r sum_{i=0}^{n-2} 2^i ] - This is a sum of a geometric series, calculated as: [ sum_{i=0}^{n-2} 2^i = 2^{n-1} - 1 ] - Thus, the total circumference of the smaller circles is: [ 2 pi r (2^{n-1} - 1) ] 8. **Meeting Analysis**: - The circumference of the largest circle ( C_{n-1} ) is: [ 2 pi cdot 2^{n-1} r ] - If one person enters ( C_{n-1} ), the second person, traveling in the opposite direction, has enough time to travel through all other circles: [ 2 pi r (2^{n-1} - 1) ] - This summed length is less than the circumference of ( C_{n-1} ), ensuring the second person will enter ( C_{n-1} ) before the first person finishes it. 9. **Conclusion**: - Given they move at the same speed and in opposite directions, their paths must intersect on the largest circle ( C_{n-1} ). - Therefore, they will inevitably meet at some point. [ boxed{} ]

answer:To prove that the sequence (a_1, a_2, ldots, a_{2015}) consists of integers, we will use mathematical induction and properties of cubic and quadratic equations. 1. **Base Case:** Consider the first equation (a_1^3 = a_1^2). This simplifies to: [ a_1^3 - a_1^2 = 0 implies a_1^2(a_1 - 1) = 0 ] Therefore, (a_1 = 0) or (a_1 = 1). 2. **Inductive Step:** Assume that for some (k), the sequence (a_1, a_2, ldots, a_k) are all integers. We need to show that (a_{k+1}) is also an integer. Consider the equation: [ a_1^3 + a_2^3 + cdots + a_{k+1}^3 = (a_1 + a_2 + cdots + a_{k+1})^2 ] By the inductive hypothesis, (a_1, a_2, ldots, a_k) are integers. Let (S_k = a_1 + a_2 + cdots + a_k) and (T_k = a_1^3 + a_2^3 + cdots + a_k^3). Then the equation becomes: [ T_k + a_{k+1}^3 = (S_k + a_{k+1})^2 ] Expanding the right-hand side, we get: [ T_k + a_{k+1}^3 = S_k^2 + 2S_k a_{k+1} + a_{k+1}^2 ] Rearranging terms, we obtain: [ a_{k+1}^3 - a_{k+1}^2 - 2S_k a_{k+1} + (T_k - S_k^2) = 0 ] This is a cubic equation in (a_{k+1}). By the Rational Root Theorem, any rational solution of this equation must be an integer divisor of the constant term (T_k - S_k^2). 3. **Analysis of Roots:** Since (a_1, a_2, ldots, a_k) are integers, (T_k) and (S_k) are also integers. Therefore, (T_k - S_k^2) is an integer. The possible rational roots of the cubic equation are integers that divide (T_k - S_k^2). We need to check the possible values of (a_{k+1}): - If (a_{k+1} = 0), the equation simplifies to (T_k = S_k^2), which is true by the inductive hypothesis. - If (a_{k+1} = -S_k), the equation simplifies to (T_k + (-S_k)^3 = 0), which is also true by the properties of sums of cubes and squares. Therefore, (a_{k+1}) must be an integer. By induction, we have shown that all terms (a_1, a_2, ldots, a_{2015}) are integers. (blacksquare)

answer:To solve this problem, we start by understanding the given conditions and what we need to find. We are given that: - Each type A truck can carry 20 tons of goods. - Each type B truck can carry 15 tons of goods. - We need to transport 300 tons of goods in total. - Exactly 7 type A trucks are used. We are asked to find the minimum number of type B trucks needed to transport all the goods without overloading any truck. Let's denote the number of type B trucks needed as x. The total carrying capacity of the trucks must be at least 300 tons. The carrying capacity contributed by type A trucks is 7 times 20 tons, and the carrying capacity contributed by type B trucks is 15x tons. Therefore, the equation representing the total carrying capacity is: [7 times 20 + 15x geq 300] Simplifying this equation: [140 + 15x geq 300] Subtracting 140 from both sides gives: [15x geq 160] Dividing both sides by 15 gives: [x geq frac{160}{15}] [x geq frac{32}{3}] Since x represents the number of trucks and must be a whole number, we round up to the nearest whole number greater than frac{32}{3}, which is 11. Therefore, at least 11 type B trucks are needed to transport all the goods. Thus, the minimum number of type B trucks required is boxed{11}.

answer:- First, treat the three lily plants as a single unit. This transforms the arrangement problem into arranging six units (five rose plants + one lily plant group). - The number of ways to arrange these six units is 6!. - Within the lily plant group, there are 3! ways to arrange the lily plants. - Therefore, the total number of arrangements is calculated by multiplying these two results: 6! times 3!. Calculation: [ 6! = 720 ] [ 3! = 6 ] [ text{Total arrangements} = 720 times 6 = 4320 ] [boxed{4320}]