MathDB

4

Part of 2010 Indonesia TST

Problems(10)

probability concerning 3n cards with a different number each

Source: 2010 Indonesia TST stage 2 test 2 p4

12/16/2020
Given 3n3n cards, each of them will be written with a number from the following sequence: 2,3,...,n,n+1,n+3,n+4,...,2n+1,2n+2,2n+4,...,3n+32, 3, ..., n, n + 1, n + 3, n + 4, ..., 2n + 1, 2n + 2, 2n + 4, ..., 3n + 3 with each number used exactly once. Then every card is arranged from left to right in random order. Determine the probability such that for every ii with 1≤i≤3n1\le i \le 3n, the number written on the ii-th card, counted from the left, is greater than or equal to ii.
probabilitycombinatorics
300 parliament members are divided into 3 chambers of 100 each

Source: 2010 Indonesia TST stage 2 test 1 p4

12/16/2020
300300 parliament members are divided into 33 chambers, each chamber consists of 100100 members. For every 22 members, they either know each other or are strangers to each other.Show that no matter how they are divided into these 33 chambers, it is always possible to choose 22 members, each from different chamber such that there exist 1717 members from the third chamber so that all of them knows these two members, or all of them are strangers to these two members.
combinatorics
Indonesian TST 2010 Problem 3 Test 3 Stage 3

Source:

12/13/2010
How many natural numbers (a,b,n)(a,b,n) with gcd(a,b)=1 gcd(a,b)=1 and n>1 n>1 such that the equation xan+ybn=22010 x^{an} +y^{bn} = 2^{2010} has natural numbers solution (x,y) (x,y)
number theory unsolvednumber theory
n! + 1 \ge \sum_{j \in J}a_j >\sqrt {n! + (n - 1)n}

Source: 2010 Indonesia TST stage 2 test 4 p4

12/16/2020
Given a positive integer nn and I={1,2,...,k}I = \{1, 2,..., k\} with kk is a positive integer. Given positive integers a1,a2,...,aka_1, a_2, ..., a_k such that for all i∈Ii \in I: 1≤ai≤n1 \le a_i \le n and ∑i=1kai≥2(n!).\sum_{i=1}^k a_i \ge 2(n!). Show that there exists J⊆IJ \subseteq I such that n!+1≥∑j∈Jaj>n!+(n−1)nn! + 1 \ge \sum_{j \in J}a_j >\sqrt {n! + (n - 1)n}
inequalitiesalgebra
exactly \sqrt[2010]{n} positive integers x \le n : p^{2010}|x^p-1, q^{2010}|x^-1

Source: 2010 Indonesia TST stage 2 test 5 p4

12/16/2020
Let nn be a positive integer with n=p2010q2010n = p^{2010}q^{2010} for two odd primes pp and qq. Show that there exist exactly n2010\sqrt[2010]{n} positive integers x≤nx \le n such that p2010∣xp−1p^{2010}|x^p - 1 and q2010∣xq−1q^{2010}|x^q - 1.
number theory
altitude, median, bisector, and trisector

Source: Indonesia IMO 2010 TST, Stage 1, Test 1, Problem 4

11/12/2009
Let ABC ABC be a non-obtuse triangle with CH CH and CM CM are the altitude and median, respectively. The angle bisector of ∠BAC \angle BAC intersects CH CH and CM CM at P P and Q Q, respectively. Assume that \angle ABP\equal{}\angle PBQ\equal{}\angle QBC, (a) prove that ABC ABC is a right-angled triangle, and (b) calculate BPCH \dfrac{BP}{CH}. Soewono, Bandung
trigonometrygeometrygeometric transformationreflectioncircumcircleangle bisectorsimilar triangles
calculation of sum of 2^(f(n))

Source: Indonesia IMO 2010 TST, Stage 1, Test 2, Problem 4

11/12/2009
For each positive integer n n, define f(n) f(n) as the number of digits 0 0 in its decimal representation. For example, f(2)\equal{}0, f(2009)\equal{}2, etc. Please, calculate S\equal{}\sum_{k\equal{}1}^{n}2^{f(k)}, for n\equal{}9,999,999,999. Yudi Satria, Jakarta
combinatorics proposedcombinatorics
inradii 3 times!

Source: Indonesia IMO 2010 TST, Stage 1, Test 3, Problem 4

11/12/2009
Let ABC ABC be an acute-angled triangle such that there exist points D,E,F D,E,F on side BC,CA,AB BC,CA,AB, respectively such that the inradii of triangle AEF,BDF,CDE AEF,BDF,CDE are all equal to r0 r_0. If the inradii of triangle DEF DEF and ABC ABC are r r and R R, respectively, prove that r\plus{}r_0\equal{}R. Soewono, Bandung
geometryrectangleincenterratiogeometry proposed
phi gcd >= gcd phi

Source: Indonesia IMO 2010 TST, Stage 1, Test 4, Problem 4

11/12/2009
Prove that for all integers m m and n n, the inequality \dfrac{\phi(\gcd(2^m \plus{} 1,2^n \plus{} 1))}{\gcd(\phi(2^m \plus{} 1),\phi(2^n \plus{} 1))} \ge \dfrac{2\gcd(m,n)}{2^{\gcd(m,n)}} holds. Nanang Susyanto, Jogjakarta
number theorygreatest common divisorinequalitiesnumber theory proposed
erasing some digits

Source: Indonesia IMO 2010 TST, Stage 1, Test 5, Problem 4

11/12/2009
Prove that the number (9999…99⏟2005)2009 (\underbrace{9999 \dots 99}_{2005}) ^{2009} can be obtained by erasing some digits of (9999…99⏟2008)2009 (\underbrace{9999 \dots 99}_{2008}) ^{2009} (both in decimal representation). Yudi Satria, Jakarta
combinatorics proposedcombinatorics