The Mother of Cases Problems

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0 replies

jlacosta

Apr 2, 2025

0 replies

Ez inequality

m4thbl3nd3r 3

N 3 minutes ago by lbh_qys

Let $a,b,c>0$ . Prove that $\sum \frac{ab^2}{a^2+2b^2+c^2}\le \frac{a+b+c}{4}$

3 replies

+1 w

m4thbl3nd3r

Yesterday at 3:57 PM

lbh_qys

3 minutes ago

Integer polynomial commutes with sum of digits

cjquines0 40

N an hour ago by ihategeo_1969

Source: 2016 IMO Shortlist N1

For any positive integer $k$ , denote the sum of digits of $k$ in its decimal representation by $S(k)$ . Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$ , the integer $P(n)$ is positive and $S(P(n)) = P(S(n)).$
Proposed by Warut Suksompong, Thailand

40 replies

cjquines0

Jul 19, 2017

ihategeo_1969

an hour ago

easy geo

ErTeeEs06 3

N an hour ago by NicoN9

Source: BxMO 2025 P3

Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ . Let $D, E, F$ be the midpoints of the arcs $\stackrel{\frown}{BC}, \stackrel{\frown}{CA}, \stackrel{\frown}{AB}$ of $\Omega$ not containing $A, B, C$ respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$ . Show that $I, D'$ and the midpoint $M$ of $EF$ lie on a line.

3 replies

ErTeeEs06

Yesterday at 11:13 AM

NicoN9

an hour ago

2024 IMO P1

EthanWYX2009 103

N an hour ago by ashwinmeena

Source: 2024 IMO P1

Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$ is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$ )

Proposed by Santiago Rodríguez, Colombia

103 replies

EthanWYX2009

Jul 16, 2024

ashwinmeena

an hour ago

No more topics!

The Mother of Cases Problems

KAME06 1

N Apr 6, 2025 by mathprodigy2011

Source: Ecuador National Olympiad OMEC level U 2024 P5 Day 2

Let $p(n)$ the product of all $n$ 's positive divisors, where $n \in \mathbb{Z}$ .
Find all $n \in \mathbb{Z}$ such that $\sqrt[2024]{p(n)} \in \mathbb{Z}$ and is not a perfect power of another integer.

1 reply

KAME06

Apr 5, 2025

mathprodigy2011

Apr 6, 2025

The Mother of Cases Problems

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number theory

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Source: Ecuador National Olympiad OMEC level U 2024 P5 Day 2

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KAME06

155 posts

KAME06

#1 h Apr 5, 2025, 10:46 PM

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This post has been edited 2 times. Last edited by KAME06, Apr 5, 2025, 10:48 PM

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mathprodigy2011

322 posts

mathprodigy2011

#2 h Apr 6, 2025, 2:29 AM

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KAME06 wrote:

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Let s(n) be the number of divisors of n for simplicity
Case 1: n is a perfect square

Let n=p^2 for any integer p. Then the product of the divisors of p^2 is just p^(s(n)) which can be verified easily. If p is not a perfect power then this requires s(n)=2024 which is impossible for any perfect square. Now we have to consider if p is a perfect power. s(n) MUST be odd since n is a perfect square so let p=q^8k for an integer q. Then n=q^16k. and s(n) needs to be 253/k. Casework on k is enough to show that there are no cases when n is a perfect square.

Quick note: if n is not a perfect square then p(n)=n^(s(n)/2) therefore we will transform our question to (n^s(n))^4048

Case 2: n is not a perfect square/power. Then p(n) = n^(s(n)/2). So any n such that there are 4048 divisors and n is not a perfect power work.

Case 3: n is an odd perfect power. Let n=p^a. Then p(n)= n^(s(n)/2). It is easy to see a must be a factor of 4048 or else if the value turns out to be an integer; it will have the prime factors of a which are not shared with 2024. So now we only have 3 more sub-cases. Let n= p^11. Then p^(s(n)) is the product. s(n) needs to be 4048/11 = 368 = 2^4*23. Some quick casework shows that no 11th power has 368 divisors. Same things happen with p^23 and p^253.

In conclusion, the set of n that satisfies this are all n with 4048 divisors that are not perfect powers.

Please tell me what i have done wrong(im pretty sure this is incorrect). If it is correct, then it is a cool problem!

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