This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 19

1985 Traian Lălescu, 2.1

Tags: floor , equation , algebra
Solve $ \quad 5\lfloor x^2\rfloor -2\lfloor x\rfloor +2=0. $

2006 Petru Moroșan-Trident, 1

Solve in the reals the equation $ 2^{\lfloor\sqrt[3]{x}\rfloor } =x. $ [i]Nedelcu Ion[/i]

2004 Nicolae Coculescu, 3

Prove the identity $ \frac{n-1}{2}=\sum_{k=1}^n \left\{ \frac{m+k-1}{n} \right\} , $ where $ n\ge 2, m $ are natural numbers, and $ \{\} $ denotes the fractional part.

2007 Nicolae Coculescu, 1

Calculate $ \left\lfloor \frac{(a^2+b^2+c^2)(a+b+c)}{a^3+b^3+c^3} \right\rfloor , $ where $ a,b,c $ are the lengths of the side of a triangle. [i]Costel Anghel[/i]

2017 Romania National Olympiad, 1

Solve in the set of real numbers the equation $ a^{[ x ]} +\log_a\{ x \} =x , $ where $ a $ is a real number from the interval $ (0,1). $ $ [] $ and $ \{\} $ [i]denote the floor, respectively, the fractional part.[/i]

2014 Indonesia MO Shortlist, A1

Let $a, b$ be positive real numbers such that there exist infinite number of natural numbers $k$ such that $\lfloor a^k \rfloor + \lfloor b^k \rfloor = \lfloor a \rfloor ^k + \lfloor b \rfloor ^k$ . Prove that $\lfloor a^{2014} \rfloor + \lfloor b^{2014} \rfloor = \lfloor a \rfloor ^{2014} + \lfloor b \rfloor ^{2014}$

2010 Laurențiu Panaitopol, Tulcea, 1

Solve in the real numbers the equation $ \arcsin x=\lfloor 2x \rfloor . $ [i]Petre Guțescu[/i]

1985 Traian Lălescu, 1.4

Let $ a $ be a non-negative real number distinct from $ 1. $ [b]a)[/b] For which positive values $ x $ the equation $$ \left\lfloor\log_a x\right\rfloor +\left\lfloor \frac{1}{3} +\log_a x\right\rfloor =\left\lfloor 2\cdot\log_a x\right\rfloor $$ is true? [b]b)[/b] Solve $ \left\lfloor\log_3 x\right\rfloor +\left\lfloor \frac{1}{3} +\log_3 x\right\rfloor =3. $

2000 District Olympiad (Hunedoara), 2

[b]a)[/b] Let $ a,b $ two non-negative integers such that $ a^2>b. $ Show that the equation $$ \left\lfloor\sqrt{x^2+2ax+b}\right\rfloor =x+a-1 $$ has an infinite number of solutions in the non-negative integers. Here, $ \lfloor\alpha\rfloor $ denotes the floor of $ \alpha. $ [b]b)[/b] Find the floor of $ m=\sqrt{2+\sqrt{2+\underbrace{\cdots}_{\text{n times}}+\sqrt{2}}} , $ where $ n $ is a natural number. Justify.

2023 Indonesia MO, 2

Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that the following equation holds for every real $x,y$: \[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor. \] [b]Note:[/b] $\lfloor x \rfloor$ denotes the greatest integer not greater than $x$.

2012 District Olympiad, 1

Solve in $ \mathbb{R} $ the equation $ [x]^5+\{ x\}^5 =x^5, $ where $ [],\{\} $ are the integer part, respectively, the fractional part.

2006 Mathematics for Its Sake, 1

Solve in the set of real numbers the equation $$ 16\{ x \}^2-8x=-1, $$ where $ \{\} $ denotes the fractional part.

2011 Romania National Olympiad, 4

Let be a natural number $ n. $ Prove that there exists a number $ k\in\{ 0,1,2,\ldots n \} $ such that the floor of $ 2^{n+k}\sqrt 2 $ is even.

India EGMO 2024 TST, 2

Tags: summation , floor , algebra
Given that $a_1, a_2, \dots, a_{10}$ are positive real numbers, determine the smallest possible value of \[\sum \limits_{i = 1}^{10} \left\lfloor \frac{7a_i}{a_i+a_{i+1}}\right\rfloor\] where we define $a_{11} = a_1$. [i]Proposed by Sutanay Bhattacharya[/i]

2014 JBMO Shortlist, 1

Solve in positive real numbers: $n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor=2014$

2012 Bogdan Stan, 3

$ \lim_{n\to\infty }\frac{1}{\sqrt[n]{n!}}\left\lfloor \log_5 \sum_{k=2}^{1+5^n} \sqrt[5^n]{k} \right\rfloor $ [i]Taclit Daniela Nadia[/i]

2011 Laurențiu Duican, 1

Solve in the real numbers the equation $ 2^{1+x} =2^{[x]} +2^{\{x\}} , $ where $ [],\{\} $ deonotes the ineger and fractional part, respectively. [i]Aurel Bârsan[/i]

2007 Nicolae Coculescu, 1

Let be two real numbers $ x,y, $ and a natural number $ n_0 $ such that $ \{ n_0x \} = \{ n_0y \} $ and $ \{ (n_0+1)x \} = \{ (n_0+1)y \} ,$ where $ \{\} $ denotes the fractional part. Show that $ \{ nx \} =\{ ny \} , $ for any natural number $ n. $ [i]Ovidiu Pop[/i]

2011 District Olympiad, 3

Let $ f:\mathbb{R}\longrightarrow\mathbb{R} $ be a function with the property that $ (f\circ f) (x) =[x], $ for any real number $ x. $ Show that there exist two distinct real numbers $ a,b $ so that $ |f(a)-f(b)|\ge |a-b|. $ $ [] $ denotes the integer part.