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Stage 6 · Number theory

10 problems · National Olympiad, first round · mathsolympiadprep.com

The answer key prints on its own page at the end.

Let $T_n$ denotes the least natural such that
$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$
Find all naturals $m$ such that $m\ge T_m$ .

Iberoamerican Olympiad

Number theory Solution and answer checking →
Let $n$ be square with 4 digits, such that all its digits are less than 6. If we add 1 to each digit the resulting number is another square. Find $n$

Cono Sur Olympiad

Number theory Solution and answer checking →
Consider the number $N=1000\ldots 0001$ consisting of the digit one followed by 2023 zeros, which are in turn followed by the digit one. How many proper divisors does $N$ have (that is, divisors strictly between 1 and $N$ ) that are also written as a digit 1 followed by some positive number of zeros, followed in turn by a digit 1?
1. A3
2. B5
3. C12
4. D14
5. E16
Italian Mathematical Olympiad

Number theory Solution and answer checking →
Solve in $\mathbb{Z}^2$ the equation: $x^2\left( 1+x^2 \right) =-1+21^y.$

Danube Mathematical Competition

Number theory Solution and answer checking →
Let $n$ to be a positive integer. Given a set $\{ a_1, a_2, \ldots, a_n \}$ of integers, where $a_i \in \{ 0, 1, 2, 3, \ldots, 2^n -1 \},$ $\forall i$ , we associate to each of its subsets the sum of its elements; particularly, the empty subset has sum of its elements equal to $0$ . If all of these sums have different remainders when divided by $2^n$ , we say that $\{ a_1, a_2, \ldots, a_n \}$ is * $n$ -complete*.
For each $n$ , find the number of * $n$ -complete* sets.

Iberoamerican Olympiad

Number theory Solution and answer checking →
Let $D_n$ be the set of divisors of $n$ . Find all natural $n$ such that it is possible to split $D_n$ into two disjoint sets $A$ and $G$ , both containing at least three elements each, such that the elements in $A$ form an arithmetic progression while the elements in $G$ form a geometric progression.

China National Olympiad

Number theory Solution and answer checking →
Let $p$ be a prime number. Prove the following theorem of Euler: the equation $p=x^{2}+3 y^{2}$ has a solution with $x, y \in \mathbb{Z}$ if and only if $p=3$ or $p \equiv 1(\bmod 3)$ . (You may use the fact that the ring of integers of $\mathbb{Q}(\sqrt{-3})$ is a principal ideal domain.)

Yau Science Award

Number theory Solution and answer checking →
Let $m>1$ be an integer. Find the smallest positive integer $n$ , such that for any integers $a_1,a_2,\ldots ,a_n; b_1,b_2,\ldots ,b_n$ there exists integers $x_1,x_2,\ldots ,x_n$ satisfying the following two conditions:
i) There exists $i\in \{1,2,\ldots ,n\}$ such that $x_i$ and $m$ are coprime
ii) $\sum^n_{i=1} a_ix_i \equiv \sum^n_{i=1} b_ix_i \equiv 0 \pmod m$

China National Olympiad

Number theory Solution and answer checking →
Positive integers $p$ , $a$ and $b$ satisfy the equation $p^2 + a^2 = b^2$ . Prove that if $p$ is a prime greater than 3, then $a$ is a multiple of 12 and $2(p + a + 1)$ is a perfect square.

BMO Round 1 2015

Number theory Solution and answer checking →
We say that a positive integer is lazy if it has no prime divisor greater than $3$ . Prove that there are at most two lazy numbers strictly between two consecutive square numbers.

KöMaL problem A 2018

Number theory Solution and answer checking →

Answer key — Stage 6 · Number theory

Worked solutions for every problem are on the site, one page per problem.

${\text{all numbers that are not powers of 2 greater than 1.}}$ open
$2025$ open
$A$ open
$(0, 0), (2, 1), (-2, 1)$ open
$2^{n(n-1)/2}$ open
$\text{No such } n \text{ exists}$ open
$\text{The equation ( p = x 2 + 3y 2 ) has a solution with ( x, y Z ) if and only if ( p = 3 ) or ( p 1 3 ).}$ open
$2\omega(m) + 1$ open
Prove it — see the worked solution open
Prove it — see the worked solution open

Problems belong to the competitions that set them. Each question on this sheet names the competition it came from, and the licences the material is used under are listed below. Free to copy for classroom use.