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

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Found problems: 10

2021 IMO Shortlist, A2

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2024 Korea Junior Math Olympiad (First Round), 14.

Find the number of positive integer $x$ that has $ {a}_{1},{a}_{2},\cdot \cdot \cdot {a}_{20} $ which follows the following ($x \ge 1000$) 1) $ {a}_{1}=2, {a}_{2}=1, {a}_{3}=x $ 2) for positive integer $n$, ($ 4 \le n \le 20 $), $ {a}_{n}={a}_{n-3}+\frac{(-2)^n}{{a}_{n-1}{a}_{n-2}} $

2022 Germany Team Selection Test, 1

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2022 Thailand TST, 1

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2022 SAFEST Olympiad, 1

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2022 Saudi Arabia IMO TST, 1

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2022 Brazil Team Selection Test, 2

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

2018 JBMO TST-Turkey, 5

Tags: NumberTheory
Let $a_1, a_2, ... , a_{1000}$ be a sequence of integers such that $a_1=3, a_2=7$ and for all $n=2, 3, ... , 999$ $a_{n+1}-a_n=4(a_1+a_2)(a_2+a_3) ... (a_{n-1}+a_n)$. Find the number of indices $1\leq n\leq 1000$ for which $a_n+2018$ is a perfect square.

2022 Germany Team Selection Test, 1

Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\] true?

1955 Miklós Schweitzer, 6

[b]6.[/b] For a prime factorisation of a positive integer $N$ let us call the exponent of a prime $p$ the integer $k$ for which $p^{k} \mid N$ but $p^{k+1} \nmid N$; let, further, the power $p^{k}$ be called the "contribution" of $p$ to $N$. Show that for any positive integer $n$ and for any primes $p$ and $q$ the contibution of $p$ to $n!$ is greater than the contribution of $q$ if and only if the exponent of $p$ is greater than that of $q$.