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Stage 7 · Number theory

10 problems · National Olympiad second round; IMO P1/P4 · mathsolympiadprep.com

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

Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n !$ in increasing order as $1=d_1<d_2<\cdots<d_k=n!$ , then we have $d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1} .$
Contents 1 Solution (Explanation of Video) 2 Solution 2 3 Video Solution 4

USAMO

Number theory Solution and answer checking →
Find all pairs $(p,q)$ of prime numbers which $p>q$ and
$\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}$
is an integer.

IMO Shortlist

Number theory Solution and answer checking →
Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ .

USAJMO

Number theory Solution and answer checking →
Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with $\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1$
*

IMO Shortlist

Number theory Solution and answer checking →
Find all positive integers $a,n\ge1$ such that for all primes $p$ dividing $a^n-1$ , there exists a positive integer $m<n$ such that $p\mid a^m-1$ .

USA Team Selection Test

Number theory Solution and answer checking →
Find all positive integers $n$ such that there exists a unique integer $a$ such that $0\leq a < n!$ with the following property:
$\text{n! a n 1}$
*

IMO Shortlist

Number theory Solution and answer checking →
For each integer $n \geq 1$ , let $f(n)$ be the number of lists of different positive integers starting with $1$ and ending with $n$ , in which each term except the last divides its successor. Prove that for each integer $N \geq 1$ there is an integer $n \geq 1$ such that $N$ divides $f(n)$ .
(So $f(1) = 1$ , $f(2) = 1$ and $f(6) = 3$ .)

BMO Round 1 2023

Number theory Solution and answer checking →
Let $p\ge 3$ be a prime number and $0\le r\le p-3$ . Let $x_1, x_2, \ldots, x_{p-1+r}$ be integer numbers satisfying $\sum_{j=1}^{p-1+r}x_j^k\equiv r~ \textrm{(mod}\ p\textrm{)}$ for all $1\le k\le p-2$ .

KöMaL problem A 2021

Number theory Solution and answer checking →
Consider the infinite sequences of positive integer numbers, in which each positive integer number to meet among elements of this sequence is equal once. Let $\{a_n\}$ , $n \ge 1$ be such a sequence. Name it sequence "consecutive" if for each positive integer number $k$ and for any positive integer numbers $n$ and $m$ such that $a_n < a_m$ , the following inequality holds: $a_{kn} < a_{km}$ . For example, the sequence $a_n = n$ is "consecutive".
a) Prove that there exists a "consecutive" sequence which is different from $a_n = n$ .
b) Does there exist a "consecutive" sequence for which the inequality $a_n \neq n$ , $n \geq 2$ holds?
c) Does there exist a "consecutive" sequence for which the inequality $a_n \neq n$ , $n \geq 1$ holds?

Problems of Ukrainian Authors

Number theory Solution and answer checking →
Let $p$ be a given prime. For each prime $r$ , we define the function as follows
$F(r) = \frac{(p^{r p} - 1)(p - 1)}{(p^r - 1)(p^p - 1)}$
1. Show that $F(r)$ is a positive integer for any prime $r \neq p$ .
2. Show that $F(r)$ and $F(s)$ are coprime for any primes $r$ and $s$ such that $r \neq p$ , $s \neq p$ and $r \neq s$ .
3. Fix a prime $r \neq p$ . Show that there is a prime divisor $q$ of $F(r)$ such that $p \mid q-1$ but $p^2 \nmid q-1$ .

Saudi Arabian Mathematical Competitions

Number theory Solution and answer checking →

Answer key — Stage 7 · Number theory

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

The integers $n \geq 3$ that satisfy the given property are $n = 3$ and $n = 4$ . open
$(3, 2)$ open
$\boxed{6097392}$ open
$\text{[ (x 1,x 2, ,x 2011 )=(1,k, ,k) with k=2+3+ +2011=2023065. ]}$ open
$(2, 6), (2^k - 1, 2), (1, n) \text{ for any } n \ge 1$ open
$\text{All prime numbers or } n = 1$ open
Prove it — see the worked solution open
Prove it — see the worked solution open
Prove it — see the worked solution open
Prove it — see the worked solution open

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