Regarding Maaths olympiad prepration

by omega2007, Apr 4, 2025, 3:13 PM

<Hey Everyone'>
I'm 10 grader student and Im starting prepration for maths olympiad..>>> From scratch (not 2+2=4 )

Do you haves compilled resources of Handouts,
PDF,
Links,
List of books topic wise

which are shared on AOPS (and from your prespective) for maths olympiad and any useful thing, which will help me in boosting Maths olympiad prepration.
L

Problem 1

by SlovEcience, Apr 4, 2025, 3:00 PM

Prove that
\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]for \( 1 \leq k \leq p-1 \), where \( C(n, m) \) is the binomial coefficient \( n \) choose \( m \).

Burak0609

by Burak0609, Apr 4, 2025, 2:41 PM

$a^7(a-1)=19b(19b+2) \implies a^7(a-1)+1=(19b+1)^2$.
So we can see $(19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1$ and $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,19$ but $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1$ because $(19b+1)^2 \equiv 0(mod 19)$. I mean $a^2-a+1$ and $a^6-a^4-a^3+a+1$ are perfect squares. $a^2 \le a^2-a+1 \le (a+1)^2$. a should be 0 or 1 because of $a^2 \le a^2-a+1 \le (a+1)^2$. We have two solution. These are $(a,b)=(0,0),(1,0)

Bashing??

by John_Mgr, Apr 4, 2025, 1:32 PM

I have learned little about what bashing mean as i am planning to start geo, feels like its less effort required and doesnt need much knowledge about the synthetic solutions?
what do you guys recommend ? also state the major difference of them... especially of bashing pros and cons..

Problem 1

by blug, Apr 4, 2025, 11:46 AM

Inspired by JK1603JK

by sqing, Apr 4, 2025, 3:31 AM

Let $ a,b,c\geq 0 $ and $ab+bc+ca=1.$ Prove that$$\frac{abc-2}{abc-1}\ge \frac{4(a^2b+b^2c+c^2a)}{a^3b+b^3c+c^3a+1} $$
This post has been edited 1 time. Last edited by sqing, Today at 8:23 AM

You'll be sure of the answer

by egxa, Dec 17, 2024, 7:47 AM

Let $n$ be a positive integer, and let $1=d_1<d_2<\dots < d_k=n$ denote all positive divisors of $n$, If the following conditions are satisfied:
$$ 2d_2+d_4+d_5=d_7$$$$ d_3 d_6 d_7=n$$$$ (d_6+d_7)^2=n+1$$
find all possible values of $n$.

An easy 3 variable equation

by BarisKoyuncu, Dec 23, 2022, 6:39 PM

For which real numbers $a$, there exist pairwise different real numbers $x, y, z$ satisfying
$$\frac{x^3+a}{y+z}=\frac{y^3+a}{x+z}=\frac{z^3+a}{x+y}= -3.$$

Solve a^7(a-1)=19b(19b+2) over Z

by BarisKoyuncu, Mar 16, 2022, 1:05 PM

Find all pairs of integers $(a,b)$ satisfying the equation $a^7(a-1)=19b(19b+2)$.

A simple power

by Rushil, Oct 16, 2005, 5:21 AM

Prove that the ten's digit of any power of 3 is even.

♪ i just hope you understand / sometimes the clothes do not make the man ♫ // https://beta.vero.site/

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