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
and train with the best! Please note that early bird pricing ends August 19th!
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.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 1, 2025
0 replies
IMO online scoreboard
Shayan-TayefehIR   124
N 3 minutes ago by ephone
Is there still an active link for IMO's online scoreboard?, I guess the scoring process is not over yet and that old link doesn't work...
124 replies
+2 w
Shayan-TayefehIR
Jul 18, 2025
ephone
3 minutes ago
IMO ShortList 1998, combinatorics theory problem 5
orl   48
N 9 minutes ago by TwentyIQ
Source: IMO ShortList 1998, combinatorics theory problem 5
In a contest, there are $m$ candidates and $n$ judges, where $n\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]
48 replies
orl
Oct 22, 2004
TwentyIQ
9 minutes ago
Find all angles
Martin.s   1
N 25 minutes ago by aidan0626
Find all angles $\theta \ne 0$ such that
$$\tan(11\theta) = \tan(111\theta) = \tan(1111\theta) = \tan(11111\theta) = \cdots$$
1 reply
Martin.s
40 minutes ago
aidan0626
25 minutes ago
Maping a circle to a polygon by a polynomial!
goldeneagle   2
N 30 minutes ago by HHGB
Source: Iran 3rd round 2013 - Algebra Exam - Problem 5
Prove that there is no polynomial $P \in \mathbb C[x]$ such that set $\left \{ P(z) \; | \; \left | z \right | =1 \right \}$ in complex plane forms a polygon. In other words, a complex polynomial can't map the unit circle to a polygon.
(30 points)
2 replies
goldeneagle
Sep 11, 2013
HHGB
30 minutes ago
No more topics!
A nice and easy gem off of StackExchange
NamelyOrange   2
N May 3, 2025 by Royal_mhyasd
Source: https://math.stackexchange.com/questions/3818796/
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.


Rephrased: Find (with proof) all solutions to $2^a5^b-2^c5^d=1$ over $(\mathbb{N}_0)^4$.
2 replies
NamelyOrange
May 2, 2025
Royal_mhyasd
May 3, 2025
A nice and easy gem off of StackExchange
G H J
G H BBookmark kLocked kLocked NReply
Source: https://math.stackexchange.com/questions/3818796/
The post below has been deleted. Click to close.
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NamelyOrange
566 posts
#1
Y by
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.


Rephrased: Find (with proof) all solutions to $2^a5^b-2^c5^d=1$ over $(\mathbb{N}_0)^4$.
This post has been edited 3 times. Last edited by NamelyOrange, May 3, 2025, 8:45 PM
Z K Y
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NamelyOrange
566 posts
#2
Y by
Bump. I rephrased the question. It's easier than it looks :D
Z K Y
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Royal_mhyasd
121 posts
#3
Y by
Let's start by noticing that either a or c has to be 0. If this isn't true, then they're both >=1, so 2^a and 2^c are even, so 1 is even which is a contradiction. Let's analyze c=0 :
2^a * 5^b - 5^d = 1
We also notice that either b or d has to be 0, since otherwise we'd have 5^b is divisible by 5 and 5^d is divisible by 5, so 1 is divisible by 5 which is false
Let's analyze d=0 :
2^a * 5^b = 2. The only solution is evidently a=1 and b=0, so we have a=1, b=0, c=0, d=0.
If b=0 :
2^a = 1 + 5^d. a=0 yields no solution, for a=1 we have d=0 which we already found in the previous case, and for a>=2 we have 2^a == 0 (mod 4), but 5^d + 1 == 2 (mod 4), contradiction.
If a=0 :
5^b - 2^c * 5^d = 1. This time we obviously can't have b=0, so d=0.
5^b - 2^c = 1, 5^b = 2^c + 1.
If c=odd we have 2^c == -1 (mod 3), so 2^c + 1 == 0 (mod 3), so 5^b is divisible by 3, contradiction. Therefore we have c=even. Let's denote c = 2x, where x is a natural number.
5^b = 4^x + 1. b=1 obviously yields x=1, so c=2, so a=0, b=1, c=2, d=0.
For b>=2 :
If x was even, we'd have 4^x == (-1)^x == 1 (mod 5), so 4^x+1 == 2 (mod 5), contradiction. Therefore x is odd.
5^b = 4^x + 1^x, 5 | 4 + 1 and x is odd, (4;5)=1, (1;5)=1 means we can use L.T.E, so v5(5^b) = v5(4+1) + v5(x), so b = 1 + v5(x), so v5(x) = b - 1
x>= 5^(b-1)
Since b>=2 we have 5^(b-1) > 2b (easy to prove through induction) so x > 2b.
Therefore, 4^x > 4^2b = 16^b, so 4^x + 1 > 16^b + 1, 5^b > 16^b + 1, contradiction.

So S = { (1, 0, 0, 0), (0, 1, 2, 0)}
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