answer:Let `x` denote the common sum on each line. Then `5x` gives the sum of all numbers `A, B, ldots, J`, with `K` counted five times. Since the sum of the digits from `1` to `10` is `1 + 2 + cdots + 10 = 55`, we have `5x = 55 + 4K`. Thus, `55 + 4K` must be a multiple of `5`. - Solve `55 + 4K = 5m` for some integer `m`. Simplifying, we get `11 + 4K/5 = m`. - `4K` must be a multiple of `5`. The possible values of `K` that satisfy this are `K in {5, 10}`. Consider `K = 5`: - Then `5x = 55 + 4*5 = 75`, so `x = 15`. - The sum of each pair of diametrically opposite vertices along with `K` is `15`. Therefore, each pair sums to `15 - 5 = 10`. - Possible pairs: {1, 9}, {2, 8}, {3, 7}, {4, 6}, plus `10` which stands alone but is only counted once and hence does not affect the pairing. There are `5! = 120` ways to assign the five pairs around `K`, and `2^4 = 16` ways to assign the numbers in each pair (excluding `10`). Hence, for `K = 5`, there are `120 * 16 = 1920` ways. For `K = 10`: - Similar calculations show also `1920` ways. Thus, the total number of ways to label the vertices is 2 cdot 1920 = boxed{3840}.

answer:1. **Define the Fibonacci Sequence:** Let {F_n} denote the Fibonacci sequence with initial conditions F_0 = 1 and F_1 = 1, and the recurrence relation F_n = F_{n-1} + F_{n-2} for n geq 2. 2. **Identify the Problem Constraints:** We need to find the maximal possible value of the sum a_1 + a_2 + cdots + a_n for a nice n-tuple (a_1, a_2, ldots, a_n), where: - gcd(a_1, a_2, ldots, a_n) = 1 - a_i mid (a_{i-1} + a_{i+1}) for all i = 1, 2, ldots, n (with a_0 = a_n and a_{n+1} = a_1). 3. **Consider the Maximal Element:** Suppose a_i is a maximal element in the n-tuple such that a_{i-1} is not maximal. Then, we must have a_i = a_{i-1} + a_{i+1}. 4. **Delete the Maximal Element:** If we delete a_i, the resulting (n-1)-tuple is still nice. This is because: - a_{i+1} equiv a_i pmod{a_{i-1}} implies a_{i-1} mid (a_{i-2} + a_i) implies a_{i-1} mid (a_{i-2} + a_{i+1}). - Similarly, a_{i+1} mid (a_{i-1} + a_{i+2}). - The gcd condition is preserved: [ 1 = gcd(a_1, ldots, a_{i-1}, a_{i-1} + a_{i+1}, a_i, ldots, a_n) = gcd(a_1, ldots, a_{i-1}, 0, a_{i+1}, ldots, a_n) = gcd(a_1, ldots, a_{i-1}, a_{i+1}, ldots, a_n). ] 5. **Base Case and Induction:** - For n = 1, the only nice 1-tuple is (1). - For n = 2, the only nice 2-tuples are (1, 2) and (1, 1) up to cyclic permutations. - Assume the claim is true for n. For an (n+1)-tuple T, the maximal element is at most F_{n+1}, and the second-largest element is at most F_n. 6. **Inductive Step:** - If the claim holds for n, then for an (n+1)-tuple T, the maximal element is the sum of two elements from the n-tuple T' obtained by deleting it. Hence, it is at most F_{n+1}. - The second-largest element in T is still in T', so it is at most F_n. 7. **Sum of Fibonacci Numbers:** By successively deleting maximal elements and applying the above claim, the maximum value of a_1 + a_2 + cdots + a_n is F_1 + F_2 + cdots + F_n. By induction, it is easy to show that this equals F_{n+2} - 2. The final answer is boxed{F_{n+2} - 2}

answer:To solve the given problem, let's break it down step by step: 1. Start with the original expression: [4sqrt{24}times dfrac{sqrt{6}}{8}div sqrt{3}-3sqrt{3}] 2. Simplify sqrt{24} as 2sqrt{6} (since 24 = 4 times 6 and sqrt{4} = 2), and then rewrite the expression: [4cdot 2sqrt{6}times dfrac{sqrt{6}}{8}div sqrt{3}-3sqrt{3}] 3. Notice that 4cdot 2 = 8, so the expression simplifies to: [8sqrt{6}times dfrac{sqrt{6}}{8}times dfrac{1}{sqrt{3}}-3sqrt{3}] 4. The 8 in the numerator and denominator cancel out, and sqrt{6}timessqrt{6} = 6, so we have: [6times dfrac{1}{sqrt{3}}-3sqrt{3}] 5. Simplify dfrac{6}{sqrt{3}} by multiplying the numerator and denominator by sqrt{3} to rationalize the denominator: [6times dfrac{sqrt{3}}{3}-3sqrt{3}] 6. Simplify the multiplication: [2sqrt{3}-3sqrt{3}] 7. Combine like terms: [2sqrt{3}-3sqrt{3} = -sqrt{3}] Therefore, the final answer is boxed{-sqrt{3}}.

answer:(1) Let A(x_{1},y_{1}), B(x_{2},y_{2}). From the system of equations begin{cases} y^{2}=2px x=my+3 end{cases}, we get y^{2}-2pmy-6p=0. Hence, y_{1}+y_{2}=2pm, y_{1}y_{2}=-6p. Also, overrightarrow{OA}cdot overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=frac{left(y_{1}y_{2}right)^{2}}{4p^{2}}+y_{1}y_{2}=9-6p=6. Solving for p, we get p=frac{1}{2}. Thus, the equation of the parabola E is y^{2}=x. (2) Proof: Since k_{1}=frac{y_{1}}{x_{1}+3}=frac{y_{1}}{my_{1}+6}, k_{2}=frac{y_{2}}{x_{2}+3}=frac{y_{2}}{my_{2}+6}, we have frac{1}{k_{1}}=m+frac{6}{y_{1}}, frac{1}{k_{2}}=m+frac{6}{y_{2}}. Therefore, frac{1}{k_{1}^{2}}+frac{1}{k_{2}^{2}}-2m^{2}=left(m+frac{6}{y_{1}}right)^{2}+left(m+frac{6}{y_{2}}right)^{2}-2m^{2} =2m^{2}+12mleft(frac{1}{y_{1}}+frac{1}{y_{2}}right)+36left(frac{1}{y_{1}^{2}}+frac{1}{y_{2}^{2}}right)-2m^{2} =2m^{2}+12mcdot frac{y_{1}+y_{2}}{y_{1}y_{2}}+36cdot frac{left(y_{1}+y_{2}right)^{2}-2y_{1}y_{2}}{y_{1}^{2}y_{2}^{2}}-2m^{2} Given y_{1}+y_{2}=2pm=m, y_{1}y_{2}=-6p=-3, we get frac{1}{k_{1}^{2}}+frac{1}{k_{2}^{2}}-2m^{2}=2m^{2}+12mtimes frac{-m}{3}+36times frac{m^{2}+6}{9}-2m^{2}=boxed{24}. Thus, frac{1}{k_{1}^{2}}+frac{1}{k_{2}^{2}}-2m^{2} is a constant.