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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
Thursday at 11:16 PM
0 replies
J
H
Interesting inequalities
sqing   0
17 minutes ago
Source: Own
Let $ a,b,c\geq 0 ,2a +ab +abc \geq 9. $ Prove that
$$a+b+c \geq 4$$$$a+b+\frac{1}{4}c \geq \frac{13}{4}$$Let $ a,b,c\geq 0 ,2a +ab +4abc \geq 9. $ Prove that
$$a+b+c+abc \geq 4$$$$a+b+4c \geq 4$$$$a+b+c \geq \frac{13}{4}$$$$a+\frac{6}{5}b+4c \geq \frac{6(\sqrt{30}-2)}{5}$$$$a+\frac{9}{5}b+4c \geq \frac{18(\sqrt 5-1)}{5}$$
0 replies
+1 w
sqing
17 minutes ago
0 replies
J
H
Hard diophant equation
MuradSafarli   3
N 23 minutes ago by MuradSafarli
Find all positive integers $x, y, z, t$ such that the equation

$$ 2017^x + 6^y + 2^z = 2025^t $$
is satisfied.
3 replies
MuradSafarli
Yesterday at 6:12 PM
MuradSafarli
23 minutes ago
J
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< NA'T = < ADT wanted, starting with a right triangle, symmetric, projections
parmenides51   3
N an hour ago by tilya_TASh
Source: JBMO Shortlist 2018 G2
Let $ABC$ be a right angled triangle with $\angle A = 90^o$ and $AD$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $AB$ and $AC$ respectively. The parallel line from $C$ to $EZ$ intersects the line $AB$ at the point $N$. Let $A' $ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. If $T$ is the point of intersection of the lines $IK$ and $DE$, prove that $\angle NA'T = \angle ADT$.
3 replies
parmenides51
Jul 22, 2019
tilya_TASh
an hour ago
J
H
x(x - y) = 8y - 7 in NxN
parmenides51   4
N an hour ago by Namisgood
Source: JBMO 2008 Shortlist N1
Find all the positive integers $x$ and $y$ that satisfy the equation $x(x - y) = 8y - 7$
4 replies
parmenides51
Oct 14, 2017
Namisgood
an hour ago
J
No more topics!
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J
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Fibonacci...?
Jackson0423   1
N Apr 30, 2025 by KAME06
The sequence \( F \) is defined by \( F_0 = F_1 = 2025 \) and for all positive integers \( n \geq 2 \), \( F_n = F_{n-1} + F_{n-2} \). Show that for every positive integer \( k \), there exists a suitable positive integer \( j \) such that \( F_j \) is a multiple of \( k \).
1 reply
Jackson0423
Apr 30, 2025
KAME06
Apr 30, 2025
J
Fibonacci...?
G H J
Reveal topic tags
number theory
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Jackson0423
65 posts
Jackson0423
#1 h  Apr 30, 2025, 4:24 PM
Y by
The sequence \( F \) is defined by \( F_0 = F_1 = 2025 \) and for all positive integers \( n \geq 2 \), \( F_n = F_{n-1} + F_{n-2} \). Show that for every positive integer \( k \), there exists a suitable positive integer \( j \) such that \( F_j \) is a multiple of \( k \).
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KAME06
158 posts
KAME06
#2 h  Apr 30, 2025, 4:50 PM
Y by
$F_0=F_1 \equiv r \pmod{k}$. We know \( F_n = F_{n-1} + F_{n-2} \), so $F_n \equiv f_nr \pmod{k}$, where $f_n$ is the $n^{th}$ term of Fibonacci sequence and of course there exist suitable positive integer $j$ such that $k \mid f_j$, so $F_j \equiv 0(r)=0 \pmod{k} \Rightarrow k \mid F_j$.
This post has been edited 1 time. Last edited by KAME06, Apr 30, 2025, 4:51 PM
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