Plan ahead for the next school year. Schedule your class today!

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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
AM-GM Problem
arcticfox009   9
N 37 minutes ago by pooh123
Let $x, y$ be positive real numbers such that $xy \geq 1$. Find the minimum value of the expression

\[ \frac{(x^2 + y)(x + y^2)}{x + y}. \]
answer confirmation
9 replies
arcticfox009
Friday at 3:01 PM
pooh123
37 minutes ago
Is it true?
lgx57   1
N 3 hours ago by alexheinis
$0<a_1,a_2\cdots ,a_n$, determine whether it is true.
$$\sum_{i=1}^n \frac{1}{a_i}\ge \sum_{i=1}^n \frac{i}{\sum_{j=1}^i a_j}$$
If not, please give a counterexample.
1 reply
lgx57
Yesterday at 3:02 PM
alexheinis
3 hours ago
Cone Sul 2020 TST 3 Brazil P2
TiagoCamara   0
4 hours ago
(Cone Sul 2020 TST 3 Brazil P2)Determine all positive integers $n$ for which $4k^2+n$ is a prime number for every $0\leq k< n$ integer.
0 replies
TiagoCamara
4 hours ago
0 replies
Limit of a sequence involving the largest odd divisor
JackMinhHieu   1
N Yesterday at 7:15 PM by mathreyes
Hi everyone,

I came across the following sequence and I’m curious about its behavior:

Let d(k) be the largest odd positive divisor of k. Define a sequence (x_n) by

x_n = (1/n) * sum_{k=1}^{n} (d(k)/k)

Question:
Does the sequence (x_n) converge? If so, what is its limit?

Any insights, proofs, or helpful observations would be appreciated. Thank you!
1 reply
JackMinhHieu
Yesterday at 5:19 PM
mathreyes
Yesterday at 7:15 PM
No more topics!
Logarithmic function
jonny   2
N May 11, 2025 by KSH31415
If $\log_{6}(15) = a$ and $\log_{12}(18)=b,$ Then $\log_{25}(24)$ in terms of $a$ and $b$
2 replies
jonny
Jul 15, 2016
KSH31415
May 11, 2025
Logarithmic function
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jonny
3455 posts
#1 • 1 Y
Y by Adventure10
If $\log_{6}(15) = a$ and $\log_{12}(18)=b,$ Then $\log_{25}(24)$ in terms of $a$ and $b$
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Mathzeus1024
1055 posts
#2
Y by
We have:

$a = \frac{\ln(15)}{\ln(6)} = \frac{\ln(3)+\ln(5)}{\ln(3)+\ln(2)}$ (i);

$b = \frac{\ln(18)}{\ln(12)} = \frac{2\ln(3)+\ln(2)}{\ln(3)+2\ln(2)}$ (ii).

If $\frac{\ln(24)}{\ln(25)} = \frac{3\ln(2)+\ln(3)}{2\ln(5)}$, then we obtain:

$\frac{\ln(24)}{\ln(25)}= \frac{\ln(3)+2\ln(2)+\ln(2)}{2\ln(5)} = \frac{\frac{2\ln(3)+\ln(2)}{b}+\ln(2)}{2[a(\ln(2)+\ln(3))-\ln(3)]} = \textcolor{red}{\frac{\ln(18)+b\ln(2)}{2b[a\ln(6)-\ln(3)]}}$.
This post has been edited 1 time. Last edited by Mathzeus1024, May 11, 2025, 2:29 PM
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KSH31415
418 posts
#3
Y by
Mathzeus1024 wrote:
$\textcolor{red}{\frac{\ln(18)+b\ln(2)}{2b[a\ln(6)-\ln(3)]}}$.
I would assume that the intent of the problem is to express $\log_{25}24$ with $a$ and $b$ an no other logarithms. This is what I did. There is probably an algebra mistake in there somewhere.

Surely there is a more direct way to find this (though my answer doesn't hint at any). Can anybody find it?
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