MathDB

3

Part of 2011 China Team Selection Test

Problems(6)

Find the number of interesting numbers

Source: China TST 2011 - Quiz 1 - D1 - P3

5/19/2011
A positive integer nn is known as an interesting number if nn satisfies  {n10k}>n1010{\ \{\frac{n}{10^k}} \} > \frac{n}{10^{10}} for all k=1,2,…9k=1,2,\ldots 9. Find the number of interesting numbers.
number theory proposednumber theorycombinatorics
Minimum possible number of elements in set

Source: China TST 2011 - Quiz 1 - D3 - P2

5/19/2011
For a given integer n≥2n\ge 2, let a0,a1,…,ana_0,a_1,\ldots ,a_n be integers satisfying 0=a0<a1<…<an=2n−10=a_0<a_1<\ldots <a_n=2n-1. Find the smallest possible number of elements in the set {ai+aj∣0≤i≤j≤n}\{ a_i+a_j \mid 0\le i \le j \le n \}.
combinatorics proposedcombinatorics
Infinitely many integers n with d(n!)-1 a composite number

Source: China TST 2011 - Quiz 2 - D1 - P3

3/24/2011
For any positive integer dd, prove there are infinitely many positive integers nn such that d(n!)−1d(n!)-1 is a composite number.
number theoryIMO Shortlistnumber theory unsolved
Find the greatest lambda so that the second inequality holds

Source: China TST 2011 - Quiz 2 - D2 - P3

5/20/2011
Let nn be a positive integer. Find the largest real number λ\lambda such that for all positive real numbers x1,x2,⋯ ,x2nx_1,x_2,\cdots,x_{2n} satisfying the inequality 12n∑i=12n(xi+2)n≥∏i=12nxi,\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\geq \prod_{i=1}^{2n} x_i, the following inequality also holds 12n∑i=12n(xi+1)n≥λ∏i=12nxi.\frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n\geq \lambda\prod_{i=1}^{2n} x_i.
inequalitiesinequalities unsolved
The minimum number of edges of G is (7n^2- 3n)/2

Source: China TST 2011 - Quiz 3 - D1 - P3

5/20/2011
Let GG be a simple graph with 3n23n^2 vertices (n≥2n\geq 2). It is known that the degree of each vertex of GG is not greater than 4n4n, there exists at least a vertex of degree one, and between any two vertices, there is a path of length ≤3\leq 3. Prove that the minimum number of edges that GG might have is equal to (7n2−3n)2\frac{(7n^2- 3n)}{2}.
algorithmcombinatorics unsolvedcombinatorics
We may perform finitely many extensions

Source: China TST 2011 - Quiz 3 - D2 - P3

5/20/2011
Let mm and nn be positive integers. A sequence of points (A0,A1,…,An)(A_0,A_1,\ldots,A_n) on the Cartesian plane is called interesting if AiA_i are all lattice points, the slopes of OA0,OA1,⋯ ,OAnOA_0,OA_1,\cdots,OA_n are strictly increasing (OO is the origin) and the area of triangle OAiAi+1OA_iA_{i+1} is equal to 12\frac{1}{2} for i=0,1,…,n−1i=0,1,\ldots,n-1. Let (B0,B1,⋯ ,Bn)(B_0,B_1,\cdots,B_n) be a sequence of points. We may insert a point BB between BiB_i and Bi+1B_{i+1} if OB→=OBi→+OBi+1→\overrightarrow{OB}=\overrightarrow{OB_i}+\overrightarrow{OB_{i+1}}, and the resulting sequence (B0,B1,…,Bi,B,Bi+1,…,Bn)(B_0,B_1,\ldots,B_i,B,B_{i+1},\ldots,B_n) is called an extension of the original sequence. Given two interesting sequences (C0,C1,…,Cn)(C_0,C_1,\ldots,C_n) and (D0,D1,…,Dm)(D_0,D_1,\ldots,D_m), prove that if C0=D0C_0=D_0 and Cn=DmC_n=D_m, then we may perform finitely many extensions on each sequence until the resulting two sequences become identical.
analytic geometrygraphing linesslopegeometryparallelogramcombinatorics unsolvedcombinatorics