answer:1. **Calculate 8 otimes 6:** [ 8 otimes 6 = frac{8 + 6}{8 - 6} = frac{14}{2} = 7 ] 2. **Calculate 2 otimes 4:** [ 2 otimes 4 = frac{2 + 4}{2 - 4} = frac{6}{-2} = -3 ] 3. **Calculate 7 otimes (-3):** [ 7 otimes (-3) = frac{7 + (-3)}{7 - (-3)} = frac{4}{10} = frac{2}{5} ] 4. **Combine the results:** Since (8 otimes 6) otimes (2 otimes 4) = 7 otimes (-3) = frac{2}{5}, the final answer is: [ frac{2{5}} ] The final answer is boxed{B}

answer:1. **Identify the given information and setup the problem:** - We are given a triangle (ABC) with ( |AC| > |BC| ). - Point (M) lies on the angle bisector of (angle C). - (BM) is perpendicular to the angle bisector of (angle C). 2. **Introduce the midpoint (N) of (BC):** - Let (N) be the midpoint of (BC). Since (N) is the midpoint, (BN = NC). 3. **Analyze the angles and parallel lines:** - Since (M) lies on the angle bisector of (angle C), (angle BCM = angle ACM). - Given that (BM) is perpendicular to the angle bisector, (angle BNM = angle C). 4. **Establish the parallelism:** - Since (angle BNM = angle C), and (N) is the midpoint of (BC), it follows that (MN parallel AC). 5. **Calculate the areas of triangles (AMC) and (ABC):** - Since (MN parallel AC) and (N) is the midpoint of (BC), triangle (BMC) is similar to triangle (AMC) with a ratio of 1:2. - The height from (M) to (AC) in (triangle AMC) is half the height from (B) to (AC) in (triangle ABC). 6. **Use the area ratio to conclude:** - The area of (triangle AMC) is half the area of (triangle ABC) because the base (AC) is the same for both triangles, and the height of (triangle AMC) is half the height of (triangle ABC). [ text{Area of } triangle AMC = frac{1}{2} times text{Area of } triangle ABC ] (blacksquare)

answer:Proof: (1) a_2=3, a_3=7, a_4=15, a_5=31... (2 points) (2) By induction, we conjecture the general formula a_{n}=2^{n}-1, ... (3 points) ① When n=1, a_{1}=1=2^{1}-1, which holds true... (4 points) ② Assume it holds for n=k, i.e., a_{k}=2^{k}-1, ... (5 points) Then for n=k+1, from a_{n+1}=2a_{n}+1 (ninmathbb{N}^+), we get: a_{k+1}=2a_{k}+1=2(2^{k}-1)+1=2^{k+1}-2+1=2^{k+1}-1... (6 points) Thus, it also holds for n=k+1; Combining ① and ②, the equation holds for all ninmathbb{N}^*, thereby proven... (7 points) (3) From (2), we know a_{100}=2^{100}-1 Since 2^{100}=(2^{4})^{25}=16^{25}=(15+1)^{25}... (8 points) Expanding: (15+1)^{25}= C_{25}^{0}15^{25}+ C_{25}^{1}15^{24}+ldots+ C_{25}^{24}15^{1}+ C_{25}^{25}15^{0}, the remainder when divided by 15 is 1,... (9 points) Therefore, a_{100}=2^{100}-1 is divisible by 15... (10 points) Thus, the final answers are: (1) a_2=3, a_3=7, a_4=15, a_5=31 boxed{text{2 points}} (2) The general formula is conjectured to be boxed{a_{n}=2^{n}-1} and proven by induction boxed{text{7 points}} (3) It is proven that boxed{a_{100}} is divisible by 15 boxed{text{10 points}}

answer:To solve this problem, we need to find the smallest positive integer ( m ) such that for any grid ( A ) and any positive integer ( k leq m ), at least one of the following conditions holds: 1. The difference between the numbers in grid ( A ) and the ( k )-th grid after ( A ) in the clockwise direction is ( k ). 2. The difference between the numbers in grid ( A ) and the ( k )-th grid after ( A ) in the anticlockwise direction is ( k ). We also need to ensure that there exists a grid ( S ) with the real number ( x ) such that for any positive integer ( k < 999 ), at least one of the following conditions holds: 1. The number in the ( k )-th grid after ( S ) in the clockwise direction is ( x + k ). 2. The number in the ( k )-th grid after ( S ) in the anticlockwise direction is ( x + k ). Let's break down the solution step by step: 1. **Define the Problem in Terms of Graph Theory:** - Consider the grids as vertices of a graph. - Connect vertices ( v_i ) and ( v_j ) with a red edge if ( a_i - a_j equiv i - j pmod{999} ). - Connect vertices ( v_i ) and ( v_j ) with a blue edge if ( a_i - a_j equiv j - i pmod{999} ). 2. **Properties of Red and Blue Edges:** - Red and blue edges form disjoint unions of cliques. - A red clique has ( a_i - i ) constant, and a blue clique has ( a_i + i ) constant. - A red and blue edge cannot exist between ( i ) and ( j ) if ( i neq j ) because this would imply ( j - i equiv a_j - a_i equiv i - j pmod{999} ), which is a contradiction. 3. **Minimum Neighbors:** - For each ( j in mathbb{Z}_{999} ) and ( 1 leq k leq m ), there must be either a red or blue edge between ( j ) and ( j+k ) or between ( j ) and ( j-k ). - Therefore, each vertex ( v_j ) has at least ( m ) red and blue neighbors. 4. **Observation and Proof:** - If ( v_{x+k_1}, v_{x-k_1}, v_{x+k_2}, v_{x-k_2} ) form a blue clique and ( k_1, k_2 < 252 ), then ( v_x ) must be part of this clique. - If ( v_x v_{x+k_1} ) is blue, we are done. Otherwise, there are two red edges from ( v_x ) to this clique, leading to a contradiction. 5. **Pigeonhole Principle:** - If there exists a vertex ( v_x ) with at least 5 red neighbors, then there exists a vertex ( v_x ) with at least 248 red neighbors among ( v_{x-251}, cdots, v_{x+251} ). 6. **Conclusion:** - If a red component has size at least 251, it must have size at least 500. - Therefore, the smallest ( m ) that satisfies the conditions is ( m = 251 ). The final answer is ( boxed{251} ).