answer:7. First, prove: if n satisfies the condition, then n not equiv 2(bmod 4).If a positive integer n satisfies the condition, let the sum of the integers marked on the three sides of each triangle be S, and the sum of the integers marked on all n-3 diagonals be m. Summing the sums of the numbers marked on the three sides of all n-2 triangles, each diagonal is counted twice, we get(n-2) S=(1+2+cdots+n)+2 m .Assume n equiv 2(bmod 4). Then (n-2) S is even.However, (1+2+cdots+n)+2 m is odd, which is a contradiction.Therefore, n neq 2(bmod 4).Next, use mathematical induction to prove: all n neq equiv 2(bmod 4) satisfy the condition.As shown in Figure 4, constructions that satisfy the condition are given for n=4,5,7.It is only necessary to prove: if n satisfies the condition, then n+4 also satisfies the condition.As shown in Figure 5, first divide the convex n+4-gon into a convex n-gon and a convex hexagon using one diagonal, and mark this diagonal with n.Since n satisfies the condition, the convex n-gon can be divided into triangles such that the sum of the numbers marked on the three sides of each triangle is equal (let this sum be S). Divide the convex hexagon into four triangles using three diagonals as shown in Figure 5, and mark the three diagonals with S-2 n-1, n+1, and S-2 n-5 respectively. It can be verified that the sum of the numbers marked on the three sides of these four triangles also equals S. Therefore, n+4 also satisfies the condition.In summary, the n that satisfy the condition are n neq 2(bmod 4).

answer:Solution: As N, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form 2^{a} cdot 5^{b}, where a=0,1,2,3,4 and b=0,1,2,3. There will be 20 of them in total, but we need to exclude 1,2,4,5,8,10 and 16.

answer:AnswerAs shown in the figure, let the cross-section be E G B_{1} F H, A_{1} F=A_{1} B_{1}=2Rightarrow C G=C E=1 Rightarrow C_{1} G=3 text {. }Then triangle F A H sim triangle G C_{1} B_{1} Rightarrow A H=frac{4}{3}, D H=frac{2}{3}.Thus, H E=frac{sqrt{13}}{3}, H F=frac{2 sqrt{13}}{3}, E F=3,so cos angle E H F=-frac{4}{13} Rightarrow sin angle E H F=frac{3 sqrt{17}}{13}Rightarrow S_{triangle E H F}=frac{1}{2} cdot frac{sqrt{13}}{3} cdot frac{2 sqrt{13}}{3} cdot frac{3 sqrt{17}}{13}=frac{sqrt{17}}{3}.Also, V_{D-E H F}=V_{F-H D E}=frac{1}{3} cdot frac{1}{2} cdot frac{2}{3} cdot 1 cdot 2=frac{2}{9},Let the distance from point D to the plane E F B_{1} be h,Therefore, V_{D-E H F}=frac{1}{3} cdot frac{sqrt{17}}{3} cdot h=frac{2}{9} Rightarrow h=frac{2}{sqrt{17}}=frac{2 sqrt{17}}{17}.

answer:When we bring the fractions to a common denominator, we then get the following numerator:(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)=a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0thus, the value of the questioned sum is 0.Number of solutions: 26.

answer:If a < 0, when x geq 0, f(x) = x - a + x = 2x - a is an increasing function, which means sufficiency is established.When a = 0, f(x) = 2|x|, which satisfies the condition of being an increasing function for x geq 0, but the condition a < 0 does not hold.Therefore, "a < 0" is a sufficient but not necessary condition for the function f(x) = |x-a| + |x| to be increasing on the interval [0, +infty).Hence, the correct choice is: boxed{text{A}}This problem mainly examines the judgment of sufficient and necessary conditions, and understanding the properties of absolute values is key to solving this problem.

answer:Keys: function change, iterates of a function.Let g(x):=2 x-f(x) for all x. We see that f circ g is the identity, which is a bijective function, implying that f and g are bijective (this is a useful classical fact). We note that g takes values in [0,1] since f(g(x)) is well-defined for all x. We denote, for n geqslant 1, g^{n}=g circ ldots circ g (composing g n times). It is easily seen by induction that g^{n}(x)=n g(x)-(n-1) x=n(g(x)-x)+x. If there exists x_{0} such that gleft(x_{0}right)-x_{0} neq 0, then for n large enough left|g^{n}left(x_{0}right)right|>1, but g takes values in [0,1] so does g^{n}, contradiction. Therefore, g(x)=x for all x, hence f(x)=x for all x. Conversely, this function is a solution.