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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Nov 1, 2024
0 replies
Problem 1: Triangle triviality
ZetaX   131
N 2 minutes ago by lnzhonglp
Source: IMO 2006, 1. day
Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.
131 replies
ZetaX
Jul 12, 2006
lnzhonglp
2 minutes ago
Problem 1 (First Day)
Valentin Vornicu   131
N 19 minutes ago by lnzhonglp
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
131 replies
1 viewing
Valentin Vornicu
Jul 12, 2004
lnzhonglp
19 minutes ago
Maximization
Butterfly   1
N 30 minutes ago by lbh_qys

Assume $a,b>0$ satisfy $a^2+b^2+\frac{1}{a}+\frac{1}{b}=\frac{27}{4}.$ Prove $2a+\frac{a}{4b}\le 5.$
1 reply
+1 w
Butterfly
an hour ago
lbh_qys
30 minutes ago
Strange Conditional Sequence
MarkBcc168   20
N 31 minutes ago by cushie27
Source: APMO 2019 P2
Let $m$ be a fixed positive integer. The infinite sequence $\{a_n\}_{n\geq 1}$ is defined in the following way: $a_1$ is a positive integer, and for every integer $n\geq 1$ we have
$$a_{n+1} = \begin{cases}a_n^2+2^m & \text{if } a_n< 2^m \\ a_n/2 &\text{if } a_n\geq 2^m\end{cases}$$For each $m$, determine all possible values of $a_1$ such that every term in the sequence is an integer.
20 replies
MarkBcc168
Jun 11, 2019
cushie27
31 minutes ago
No more topics!
5^m+6^n with all same digits
Circumcircle   5
N Nov 17, 2024 by JanHaj
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 3
Find all pairs of natural numbers $(m,n)$ such that the number $5^m+6^n$ has all same digits when written in decimal representation.
5 replies
Circumcircle
Nov 16, 2024
JanHaj
Nov 17, 2024
5^m+6^n with all same digits
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Source: Kosovo Math Olympiad 2025, Grade 11, Problem 3
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Circumcircle
64 posts
#1
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Find all pairs of natural numbers $(m,n)$ such that the number $5^m+6^n$ has all same digits when written in decimal representation.
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RagvaloD
4730 posts
#3
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$5^m+6^n$ ends with $1$, so we want to find $5^m+6^n=11...1$
Easy to prove that $5^m$ ends with $25$ for $m \geq 2$ and $6^n$ ends with $a6$ where $a$ is odd for $n\geq 2$
So for $n \geq 2$ we have that $5^m+6^n$ ends with $b1$ where $b$ is even, so $b \neq 1$
So $n=1$
$5^m+6=11...1$
But for $m \geq 2$ we have that $5^m+6$ ends with $31$

So only $(1,1)$ is solution
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Carius_MrPenguin
6 posts
#4
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Just notice that $5^m+6^n$ ends with 1, so all the digits must be 1.
for $m>1$ we have $5^m\equiv 25 \pmod{100}$ (well known) and ....25+....6=1...111, is neccesary to find a power of six with the propierty:
$100k+86=6^n$, then $6\equiv 2 \pmod{4}$ wich is only possible if $n=1$ contradiction since $6^n$ has at least 2 digits in this case.
So $m=1$, for $n>1$ we have $5+...6=11...1$, then the power of six must be $5+...96=1...11$ and again $5+...996=1...111$ infinitely times number $9$ is in its decimal representation (absurdum) so $6^n$ has only one digit and $n=1$
$m=1$ and $n=1$ is the only posibility.
This post has been edited 1 time. Last edited by Carius_MrPenguin, Nov 17, 2024, 1:25 AM
Reason: A little mistake with the $
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grupyorum
1366 posts
#5
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Note that all digits must be 1. So, $5^m+6^n\equiv 11\equiv 3\pmod{4}$. This yields $n=1$ immediately. Next, $5^m+6^n\equiv 11\pmod{25}$. If $m\ge 2$, then $5^m+6^n\equiv 6\pmod{25}$, a contradiction. So $m=n=1$.
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SimplisticFormulas
23 posts
#6
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We claim that only such pair $(m,n) $ is $(1,1)$.
Note that $5^m+6^n \equiv 5+6\equiv 1 \mod 10$, so the last digit is 1 and hence all the digits, including the second last digit is $1$. Now, taking modulo $100$, we get
$5^m \equiv 5,25 \mod 100$
$\implies 6^n \equiv 11-5,11-25 \equiv 6,86 \mod 100$
But $6^n \equiv 6,16,36,56,76,96 \mod 100$,
Hence $5^m \equiv 5 \mod 100$ and $6^n \equiv 6\mod 100$, giving is $(m,n)=(1,1)$ as the only solution.
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JanHaj
29 posts
#7 • 1 Y
Y by ItsBesi
Another beautiful problem. Here's my solution from the contest.
We will prove that the only pair is $(m,n)=(1,1)$.
Note that since $5^m+6^n \equiv 5+6 \equiv 1  \pmod {10}$ so by the problem condition all digits must be $1$.
Thus we can write:
$$ 5^m+6^n= \frac{10^k-1}{9} \iff 9(5^m+6^n)= 10^k-1 $$If $k,n \geq 3 $ then clearly $8\mid 6^n$ and $8 \mid 10^k$ by taking the last equation $\pmod 8$ we have:
$$ 5^m \equiv 9(5^m+6^n)  \equiv 10^k-1 \equiv  -1 \pmod 8$$But the equation $5^m\equiv -1 \pmod 8$ clearly has no solutions, so $k,n \leq 2$. Checking these cases we see that only the pair $(m,n)=(1,1)$ satisfies the equation. $\blacksquare$.
This post has been edited 1 time. Last edited by JanHaj, Nov 17, 2024, 3:38 PM
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