MathDB

6

Part of 2019 Iran Team Selection Test

Problems(3)

Iran TST

Source: Iranian TST 2019, first exam day 2, problem 6

6/24/2019
{an}n≥0\{a_{n}\}_{n\geq 0} and {bn}n≥0\{b_{n}\}_{n\geq 0} are two sequences of positive integers that ai,bi∈{0,1,2,⋯ ,9}a_{i},b_{i}\in \{0,1,2,\cdots,9\}. There is an integer number MM such that an,bn≠0a_{n},b_{n}\neq 0 for all n≥Mn\geq M and for each n≥0n\geq 0 (an⋯a1a0‾)2+999∣(bn⋯b1b0‾)2+999(\overline{a_{n}\cdots a_{1}a_{0}})^{2}+999 \mid(\overline{b_{n}\cdots b_{1}b_{0}})^{2}+999 prove that an=bna_{n}=b_{n} for n≥0n\geq 0.\\ (Note that (xnxn−1…x0‾)=10n×xn+⋯+10×x1+x0(\overline{x_nx_{n-1}\dots x_0}) = 10^n\times x_n + \dots + 10\times x_1 + x_0.)
Proposed by Yahya Motevassel
number theory
Iran inequality

Source: Iranian TST 2019, third exam day 2, problem 6

4/15/2019
x,yx,y and zz are real numbers such that x+y+z=xy+yz+zxx+y+z=xy+yz+zx. Prove that xx4+x2+1+yy4+y2+1+zz4+z2+1≥−13.\frac{x}{\sqrt{x^4+x^2+1}}+\frac{y}{\sqrt{y^4+y^2+1}}+\frac{z}{\sqrt{z^4+z^2+1}}\geq \frac{-1}{\sqrt{3}}.
Proposed by Navid Safaei
inequalities
Iran TST

Source: Iranian TST 2019, second exam day 2, problem 6

6/24/2019
For any positive integer nn, define the subset SnS_n of natural numbers as follow Sn={x2+ny2:x,y∈Z}. S_n = \left\{x^2+ny^2 : x,y \in \mathbb{Z} \right\}. Find all positive integers nn such that there exists an element of SnS_n which doesn't belong to any of the sets S1,S2,…,Sn−1S_1, S_2,\dots,S_{n-1}.
Proposed by Yahya Motevassel
number theory