Good divisors and special numbers.

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Good divisors and special numbers.

G H J

Reveal topic tags

AZE Al-Khwarizmi IJMO TST

L

G H BBookmark kLocked kLocked NReply

Source: Azerbaijan Al-Khwarizmi IJMO TST 2024

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

83 posts

#1 h Apr 29, 2025, 4:52 PM

Y by

$N$ is a positive integer. Call all positive divisors of $N$ which are different from $1$ and $N$ beautiful divisors.We call $N$ a special number when it has at least $2$ beautiful divisors and difference of any $2$ beautiful divisors divides $N$ as well. Find all special numbers.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1335 posts

#2 h Apr 29, 2025, 10:17 PM • 2 Y

Y by RobertRogo, Nuran2010

Nice problem!

If $N$ is composite with at least two prime divisors, we may write $N = ab$ for coprime $1 < a < b < N$ . Then $a-b$ divides $ab$ but also their gcd is $1$ (a common divisor dividing $a-b$ and $b$ also divides $a$ ), so $b-a = 1$ and $N = a(a+1)$ . If $a=2$ , then $N=6$ which works. If $a>2$ , then (as $N$ is even as a product of two consecutive integers) $a-2$ must divide $a(a+1) \equiv 2 \cdot 3 = 6$ , so $a=3,4,5,8$ , corresponding to $N = 12, 20, 30, 72$ . However, $N=20$ does not work with $2$ and $5$ , $30$ does not work with $2$ and $6$ , $72$ does not work with $2$ and $9$ , while $N=12$ works since all diiferences are $1,2,3,4$ .

Now suppose $N = p^k$ for some prime $p$ , where $k\geq 3$ from the problem statement. Then $p^2 - p$ must divide $p^k$ , so $p-1$ divides $p^{k-1} \equiv 1$ , which implies $p=2$ . Now $N = 8$ works (as $4-2=2$ ), while $k\geq 4$ does not work with $8 - 2 = 6$ .

Therefore all working $N$ are $6, 8, 12$ .

This post has been edited 1 time. Last edited by Assassino9931, Apr 29, 2025, 10:17 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

570 posts

#4 h Apr 30, 2025, 12:38 AM

Y by

May Olympiad 2021 Level 2 Problem 2

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

83 posts

#5 h Apr 30, 2025, 6:28 PM • 1 Y

Y by FarrukhBurzu

Assassino9931 wrote:

Nice problem!

If $N$ is composite with at least two prime divisors, we may write $N = ab$ for coprime $1 < a < b < N$ . Then $a-b$ divides $ab$ but also their gcd is $1$ (a common divisor dividing $a-b$ and $b$ also divides $a$ ), so $b-a = 1$ and $N = a(a+1)$ . If $a=2$ , then $N=6$ which works. If $a>2$ , then (as $N$ is even as a product of two consecutive integers) $a-2$ must divide $a(a+1) \equiv 2 \cdot 3 = 6$ , so $a=3,4,5,8$ , corresponding to $N = 12, 20, 30, 72$ . However, $N=20$ does not work with $2$ and $5$ , $30$ does not work with $2$ and $6$ , $72$ does not work with $2$ and $9$ , while $N=12$ works since all diiferences are $1,2,3,4$ .

Now suppose $N = p^k$ for some prime $p$ , where $k\geq 3$ from the problem statement. Then $p^2 - p$ must divide $p^k$ , so $p-1$ divides $p^{k-1} \equiv 1$ , which implies $p=2$ . Now $N = 8$ works (as $4-2=2$ ), while $k\geq 4$ does not work with $8 - 2 = 6$ .

Therefore all working $N$ are $6, 8, 12$ .

My solution in the exam was very similar to this one

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a