MathDB

4

Part of 2011 District Olympiad

Problems(6)

{a}+{1/a} =1 implies {a^n} +{1/a^n} =1

Source: Romanian District Olympiad 2011, Grade IX, Problem 4

10/8/2018
Let be a nonzero real number a, a, and a natural number n. n. Prove the implication: {a}+{1a}=1  ⟹  {an}+{1an}=1, \{ a \} +\left\{\frac{1}{a}\right\} =1 \implies \{ a^n \} +\left\{\frac{1}{a^n}\right\} =1 , where {} \{\} is the fractional part.
algebrafractional part
sum of elements of set { n/2+m/5 | m, n = 0, 1, 2,..., 100}

Source: 2011 Romania District VII p4

9/1/2024
Find the sum of the elements of the set M={n2+m5  ∣m,n=0,1,2,...,100}M = \left\{ \frac{n}{2}+\frac{m}{5} \,\, | m, n = 0, 1, 2,..., 100\right\}
algebranumber theory
{ \sqrt{m} } = { \sqrt{m + 2011} }

Source: 2011 Romania District VIII p4

9/1/2024
Find all positive integers mm such that {m}={m+2011}.\{\sqrt{m}\} = \{\sqrt{m+ 2011}\}.
algebraInteger Part
logarithm of logarithm of number

Source: Romanian District Olympiad 2011, Grade X, Problem 4

10/8/2018
a) Show that , if a,b>1 a,b>1 are two distinct real numbers, then log⁡alog⁡ab>log⁡blog⁡ab. \log_a\log_a b >\log_b\log_a b.
b) Show that if a1>a2>⋯>an>1 a_1>a_2>\cdots >a_n>1 are n≥2 n\ge 2 real numbers, then log⁡a1log⁡a1a2+log⁡a2log⁡a2a3+⋯+log⁡an−1log⁡an−1an+log⁡anlog⁡ana1>0. \log_{a_1}\log_{a_1} a_2 +\log_{a_2}\log_{a_2} a_3 +\cdots +\log_{a_{n-1}}\log_{a_{n-1}} a_n +\log_{a_n}\log_{a_n} a_1 >0.
logarithmsalgebraa implies b
Romania District Olympiad 2011 - Grade XI

Source:

3/12/2011
Find all the functions f:[0,1]→Rf:[0,1]\rightarrow \mathbb{R} for which we have:
∣x−y∣2≤∣f(x)−f(y)∣≤∣x−y∣,|x-y|^2\le |f(x)-f(y)|\le |x-y|,
for all x,y∈[0,1]x,y\in [0,1].
functionanalytic geometrygraphing linesslopeinequalitiestriangle inequalityreal analysis
nilpotent+unit = unit; sufficient conditions to determine the nilpotent elements

Source: Romanian District Olympiad 2011, Grade XII, Problem 4

10/8/2018
Let be a ring A. A. Denote with N(A) N(A) the subset of all nilpotent elements of A, A, with Z(A) Z(A) the center of A, A, and with U(A) U(A) the units of A. A. Prove:
a) Z(A)=A  ⟹  N(A)+U(A)=U(A). Z(A)=A\implies N(A)+U(A)=U(A) . b) card(A)∈N∧a+U(A)⊂U(A)  ⟹  a∈N(A). \text{card} (A)\in\mathbb{N}\wedge a+U(A)\subset U(A)\implies a\in N(A) .
group theoryabstract algebraRing Theorynilpotencecardinalsuperior algebra