MathDB

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Part of 2013 Iran MO (3rd Round)

Problems(5)

Bertrand's Theorem in Polynomials!!

Source: Iran 3rd round 2013 - Algebra Exam - Problem 1

9/11/2013
Let a0,a1,…,an∈Na_0,a_1,\dots,a_n \in \mathbb N. Prove that there exist positive integers b0,b1,…,bnb_0,b_1,\dots,b_n such that for 0≤i≤n:ai≤bi≤2ai0 \leq i \leq n : a_i \leq b_i \leq 2a_i and polynomial P(x)=b0+b1x+⋯+bnxnP(x) = b_0 + b_1 x + \dots + b_n x^n is irreducible over Q[x]\mathbb Q[x]. (10 points)
algebrapolynomialalgebra proposed
Easy one!

Source: Iran 3rd round 2013 - Number Theory Exam - Problem 1

9/11/2013
Let pp a prime number and dd a divisor of p−1p-1. Find the product of elements in Zp\mathbb Z_p with order dd. (mod  p\mod p). (10 points)
modular arithmeticnumber theoryrelatively primenumber theory proposedModular inverseMultiplicative order
Parallel Line Makes Tanget Circumcircle

Source: Iran Third Round 2013 - Geometry Exam - Problem 1

9/5/2013
Let ABCDEABCDE be a pentagon inscribe in a circle (O)(O). Let BE∩AD=T BE \cap AD = T. Suppose the parallel line with CDCD which passes through TT which cut AB,CEAB,CE at X,YX,Y. If ω\omega be the circumcircle of triangle AXYAXY then prove that ω\omega is tangent to (O)(O).
geometrycircumcirclegeometric transformationhomothetygeometry unsolved
Generating function and partitions

Source: Iran 3rd round 2013- Combinatorics exam problem 1

9/18/2014
Assume that the following generating function equation is correct, prove the following statement: Πi=1∞(1+x3i)Πj=1∞(1−x6j+3)=1\Pi_{i=1}^{\infty} (1+x^{3i})\Pi_{j=1}^{\infty} (1-x^{6j+3})=1 Statement: The number of partitions of nn to numbers not of the form 6k+16k+1 or 6k−16k-1 is equal to the number of partitions of nn in which each summand appears at least twice. (10 points) Proposed by Morteza Saghafian
functioncombinatorics unsolvedcombinatorics
Polystick

Source: Iran 3rd round 2013 - final exam problem 1

9/25/2014
An nn-stick is a connected figure consisting of nn matches of length 11 which are placed horizontally or vertically and no two touch each other at points other than their ends. Two shapes that can be transformed into each other by moving, rotating or flipping are considered the same. An nn-mino is a shape which is built by connecting nn squares of side length 1 on their sides such that there's a path on the squares between each two squares of the nn-mino. Let SnS_n be the number of nn-sticks and MnM_n the number of nn-minos, e.g. S3=5S_3=5 And M3=2M_3=2. (a) Prove that for any natural nn, Sn≥Mn+1S_n \geq M_{n+1}. (b) Prove that for large enough nn we have (2.4)n≤Sn≤(16)n(2.4)^n \leq S_n \leq (16)^n. A grid segment is a segment on the plane of length 1 which it's both ends are integer points. A polystick is called wise if using it and it's rotations or flips we can cover all grid segments without overlapping, otherwise it's called unwise. (c) Prove that there are at least 2n−62^{n-6} different unwise nn-sticks. (d) Prove that any polystick which is in form of a path only going up and right is wise. (e) Extra points: Prove that for large enough nn we have 3n≤Sn≤12n3^n \leq S_n \leq 12^n
Time allowed for this exam was 2 hours.
geometrygeometric transformationrotationcombinatorics unsolvedcombinatorics