MathDB

6

Part of 2014 Iran MO (3rd Round)

Problems(2)

Prove that there are 100 natural number

Source: Iranian 3rd round Number Theory exam P6

9/22/2014
Prove that there are 100 natural number a1<a2<...<a99<a100a_1 < a_2 < ... < a_{99} < a_{100} ( ai<106 a_i < 10^6) such that A , A+A , 2A , A+2A , 2A + 2A are five sets apart ?
A={a1,a2,...,a99,a100}A = \{a_1 , a_2 ,... , a_{99} ,a_{100}\}
2A={2ai∣1≤i≤100}2A = \{2a_i \vert 1\leq i\leq 100\}
A+A={ai+aj∣1≤i<j≤100}A+A = \{a_i + a_j \vert 1\leq i<j\leq 100\}
A+2A={ai+2aj∣1≤i,j≤100}A + 2A = \{a_i + 2a_j \vert 1\leq i,j\leq 100\}
2A+2A={2ai+2aj∣1≤i<j≤100}2A + 2A = \{2a_i + 2a_j \vert 1\leq i<j\leq 100\}
(20 ponits )
modular arithmeticarithmetic seriesnumber theory proposednumber theory
Polynomial of a Function

Source: Iran 3rd round 2014 - final exam problem 6

9/16/2014
PP is a monic polynomial of odd degree greater than one such that there exists a function f:R→Nf : \mathbb{R} \rightarrow \mathbb{N} such that for each x∈Rx \in \mathbb{R} ,f(P(x))=P(f(x))f(P(x))=P(f(x)) (a) Prove that there are a finite number of natural numbers in range of ff. (b) Prove that if ff is not constant then the equation P(x)−x=0P(x)-x=0 has at least two real solutions. (c) For each natural n>1n>1 prove that there exists a function f:R→Nf : \mathbb{R} \rightarrow \mathbb{N} and a monic polynomial of odd degree greater than one PP such that for each x∈Rx \in \mathbb{R} ,f(P(x))=P(f(x))f(P(x))=P(f(x)) and range of ff contains exactly nn different numbers.
Time allowed for this problem was 105 minutes.
algebrapolynomialfunctionfloor functionalgebra unsolved