Olympiad problems

2025 Romania National Olympiad, 3

Let $n \geq 2$ be a positive integer. Consider the following equation: \[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\] a) For $n=2$, solve the given equation in $\mathbb{R}$. b) Prove that, for any $n \geq 2$, the equation has at most $2$ real solutions.

2024 India Regional Mathematical Olympiad, 2

Tags: bounding , number theory

For a positive integer $n$, let $R(n)$ be the sum of the remainders when $n$ is divided by $1,2, \cdots , n$. For example, $R(4) = 0 + 0 + 1 + 0 = 1,$ $R(7) = 0 + 1 + 1 + 3 + 2 + 1 + 0 = 8$. Find all positive integers such that $R(n) = n-1$.

2023 OMpD, 2

Tags: bounding , arithmetic progression , floor function , algebra

Find all pairs $(a,b)$ of real numbers such that $\lfloor an + b \rfloor$ is a perfect square, for all positive integer $n$.

2025 Bangladesh Mathematical Olympiad, P4

Tags: number theory , prime number , diophantine equation , bounding , modular arithmetic

Find all prime numbers $p, q$ such that$$p(p+1)(p^2+1) = q^2(q^2+q+1) + 2025.$$ [i]Proposed by Md. Fuad Al Alam[/i]

2010 Contests, 1

Tags: symmetry , subset , bounding , combinatorics

A finite set of integers is called [i]bad[/i] if its elements add up to $2010$. A finite set of integers is a [i]Benelux-set[/i] if none of its subsets is bad. Determine the smallest positive integer $n$ such that the set $\{502, 503, 504, . . . , 2009\}$ can be partitioned into $n$ Benelux-sets. (A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.) [i](2nd Benelux Mathematical Olympiad 2010, Problem 1)[/i]

2019 LIMIT Category B, Problem 11

Tags: floor function , algebra , function , bounding

$$\left\lfloor\left(1\cdot2+2\cdot2^2+\ldots+100\cdot2^{100}\right)\cdot9^{-901}\right\rfloor=?$$

2019 Philippine TST, 3

Tags: divisor , bounding , infinite sequence , number theory

Let $a_1, a_2, a_3,\ldots$ be an infinite sequence of positive integers such that $a_2 \ne 2a_1$, and for all positive integers $m$ and $n$, the sum $m + n$ is a divisor of $a_m + a_n$. Prove that there exists an integer $M$ such that for all $n > M$, we have $a_n \ge n^3$.

2016 Balkan MO, 1

Tags: function , bounding , size

Find all injective functions $f: \mathbb R \rightarrow \mathbb R$ such that for every real number $x$ and every positive integer $n$,$$ \left|\sum_{i=1}^n i\left(f(x+i+1)-f(f(x+i))\right)\right|<2016$$ [i](Macedonia)[/i]

2010 Benelux, 1

Tags: symmetry , subset , bounding , combinatorics

A finite set of integers is called [i]bad[/i] if its elements add up to $2010$. A finite set of integers is a [i]Benelux-set[/i] if none of its subsets is bad. Determine the smallest positive integer $n$ such that the set $\{502, 503, 504, . . . , 2009\}$ can be partitioned into $n$ Benelux-sets. (A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.) [i](2nd Benelux Mathematical Olympiad 2010, Problem 1)[/i]

2020 Indonesia MO, 8

Tags: construction , bounding , bait , number theory

Determine the smallest natural number $n > 2$, or show that no such natural numbers $n$ exists, that satisfy the following condition: There exists natural numbers $a_1, a_2, \dots, a_n$ such that \[ \gcd(a_1, a_2, \dots, a_n) = \sum_{k = 1}^{n - 1} \underbrace{\left( \frac{1}{\gcd(a_k, a_{k + 1})} + \frac{1}{\gcd(a_k, a_{k + 2})} + \dots + \frac{1}{\gcd(a_k, a_n)} \right)}_{n - k \ \text{terms}} \]

2016 Balkan MO Shortlist, A3

Tags: function , bounding , size

Find all injective functions $f: \mathbb R \rightarrow \mathbb R$ such that for every real number $x$ and every positive integer $n$,$$ \left|\sum_{i=1}^n i\left(f(x+i+1)-f(f(x+i))\right)\right|<2016$$ [i](Macedonia)[/i]

1995 Italy TST, 3

Tags: function , inequalities , functional equation , algebra , bounding

A function $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfies the conditions \[\begin{cases}f(x+24)\le f(x)+24\\ f(x+77)\ge f(x)+77\end{cases}\quad\text{for all}\ x\in\mathbb{R}\] Prove that $f(x+1)=f(x)+1$ for all real $x$.