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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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.

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0 replies
jlacosta
May 1, 2025
0 replies
Complex number
ronitdeb   0
22 minutes ago
Let $z_1, ... ,z_5$ be vertices of regular pentagon inscribed in a circle whose radius is $2$ and center is at $6+i8$. Find all possible values of $z_1^2+z_2^2+...+z_5^2$
0 replies
ronitdeb
22 minutes ago
0 replies
Elementary Problems Compilation
Saucepan_man02   29
N 28 minutes ago by Electrodynamix777
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
29 replies
Saucepan_man02
May 26, 2025
Electrodynamix777
28 minutes ago
Cup of Combinatorics
M11100111001Y1R   5
N 37 minutes ago by MathematicalArceus
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
5 replies
M11100111001Y1R
May 27, 2025
MathematicalArceus
37 minutes ago
Generic Real-valued FE
lucas3617   4
N 39 minutes ago by GreekIdiot
$f: \mathbb{R} -> \mathbb{R}$, find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$,$y \in \mathbb{R}$
4 replies
lucas3617
Apr 25, 2025
GreekIdiot
39 minutes ago
No more topics!
All prime factors under 8
qwedsazxc   24
N May 27, 2025 by Assassino9931
Source: 2023 KMO Final Round Day 2 Problem 4
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
24 replies
qwedsazxc
Mar 26, 2023
Assassino9931
May 27, 2025
All prime factors under 8
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Source: 2023 KMO Final Round Day 2 Problem 4
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qwedsazxc
167 posts
#1 • 1 Y
Y by DEKT
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
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seojun8978
73 posts
#2
Y by
I took the test and I solved this problem. It needs some long calculations but they are not that hard. The answer seems to be 1,2,3,4,6
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Tintarn
9045 posts
#3 • 4 Y
Y by seojun8978, rightways, ljwn357, Assassino9931
Solution
This post has been edited 1 time. Last edited by Tintarn, Mar 26, 2023, 7:00 AM
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seojun8978
73 posts
#4 • 1 Y
Y by Tintarn
Tintarn wrote:
Solution
Nice Solution!! It's like 20 times shorter than my solution.
But.. Do you mean except?(expect)
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qwedsazxc
167 posts
#5
Y by
This was the easiest one on the test, probably everyone solved this.
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seojun8978
73 posts
#6 • 1 Y
Y by GuvercinciHoca
qwedsazxc wrote:
This was the easiest one on the test, probably everyone solved this.

Yes it's true. This would have been easier than geometry if there was one.
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Seungjun_Lee
526 posts
#7
Y by
seojun8978 wrote:
qwedsazxc wrote:
This was the easiest one on the test, probably everyone solved this.

Yes it's true. This would have been easier than geometry if there was one.

I think the geometry problem was as easy as this and maybe a little easier(on day 1)
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pokmui9909
185 posts
#8
Y by
I'm very curious about the prize cut. I heard there are a lot of participants who solved $4, 5$ problems.
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Seungjun_Lee
526 posts
#9
Y by
pokmui9909 wrote:
I'm very curious about the prize cut. I heard there are a lot of participants who solved $4, 5$ problems.

I did not take the test but it seems that 2 problems are given
Maybe really perfect 2problems or just 3problems will be honorable mention award
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IAmTheHazard
5005 posts
#10 • 2 Y
Y by megarnie, centslordm
Trivial by Zsigmondy
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Tintarn
9045 posts
#11
Y by
IAmTheHazard wrote:
Trivial by Zsigmondy
I agree that the problem is trivial and my first thought when I saw the problem was also Zsigmondy, but now I am quite certain that there is no detailed solution with Zsigmondy that is shorter than my elementary one in #3. (After all, there is no way around checking the small cases.)
This post has been edited 1 time. Last edited by Tintarn, Mar 26, 2023, 5:14 PM
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rightways
868 posts
#12
Y by
$Lemma:$

If $n>1$, then the greatest prime divisor of $2^n-1$ is greater than the greatest prime divisor of $n$.

Proof:

Let $p|n$ is greatest prime divisor of $n$, then for some prime $q$:
$q|2^p-1|2^n-1$, then $q>p$ because $p=ord_2(q)$. But $q$ is also a divisor of $2^n-1$, so it is not greater than gpd of $2^n-1$.

Now, in problem, by lemma we have that $7$ is greater than any prime divisor of $n$ , so we can write $n=2^a3^b5^c$ and we can easily finish problem by proving this $c<1$ and $b<2$ and $a<3$
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rightways
868 posts
#13
Y by
Can you post KJMO day 2 problems?
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
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qwedsazxc
167 posts
#14
Y by
rightways wrote:
Can you post KJMO day 2 problems?
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$

There isn't a KJMO day 2. Final KMO contestants are picked from the people who excelled KMO 2nd round or KJMO 2nd round. I had the chance to take the Final KMO because I got a silver award from the KJMO I took on November 2022.
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hectorleo123
347 posts
#15
Y by
qwedsazxc wrote:
Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
$2^2-1=2$
$2^3-1=7$
$2^4-1=3\times 5$
$gcd(2,1)=1$
By Zsigmondy's Theorem:
$\Rightarrow \exists$ prime $p \neq 2,3,5,7 / p|2^n-1, \forall n>4$
$\Rightarrow n\le 4$
But there is an exception $n=6$
$\Rightarrow n=1,2,3,4$ and $6$ are the only values that satisfy the condition $_\blacksquare$
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Aiden-1089
302 posts
#16
Y by
It is easy to check that $n=1,2,3,4,6$ works. We claim that these are the only solutions.
Clearly $2 \nmid 2^n-1$. By Zsigmondy's theorem, any other $n$ would lead to $2^n-1$ having a prime divisor $p$ where $p \neq 3$ (because $3 \mid 2^2-1$), $5$ (because $5 \mid 2^4-1$), $7$ (because $7 \mid 2^3-1$). Since $p>7$, $n$ does not satisfy the condition.
This post has been edited 1 time. Last edited by Aiden-1089, Mar 26, 2024, 4:32 AM
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cursed_tangent1434
650 posts
#17 • 1 Y
Y by GeoKing
Me when this actually appeared on our National Olympiad. We claim that the only answers are $2,3,4$ and $6$. It is easy to see that these solutions indeed work. Now, we show that there are no other solutions.

We use the well known lemma that for all $d\mid n$ for positive integers $n$,
\[a^d - b^d \mid a^n - b^n\]for all positive integers $a,b$ extensively through out this solution. We first constrict the divisors of $n$.

Claim : There exists no prime divisor $p >3$ of $n$.
Proof : Say there exists such a prime factor $p$ of $n$. Then,
\[2^p -1 \mid 2^n-1\]and since $2^n-1$ has no prime factor larger than 7, this implies that $2^p-1$ also satisfies the same property. But, it is easy to see that $2 \nmid 2^n-1$ for any $n$, $3\mid 2^n-1$ only for even $n$, $5\mid 2^n-1$ only for $4\mid n$ and $7\mid 2^n-1$ only for $3\mid n$. Since $p>3$, this means that none of these prime factors can divide $2^p-1$ implying that it has a prime factor larger than 7, which is a clear contradiction. This proves the claim.

Now, we also note that,
\[2^8-1 = 255=3 \times 5 \times 17\]so, $8 \nmid n$ (since then $17 \mid 2^n-1$), and also
\[2^9 -1 = 511 = 7 \times 73\]so $9 \nmid n$ as well. This implies that $12 \mid n$ so $n \in \{1,2,3,4,6,12\}$. Now, we can simply check all these possibilities and see that they all work except for $1$ (which we exclude for conventional reasons) and $12$ (which we exclude since $2^{12}-1= 3^2 \times 5 \times 7 \times 13$ so $13$ is a prime factor), which implies that the solution set is indeed as claimed.
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kub-inst
31 posts
#18
Y by
If $n$ satisfies the limit, then we can let $2^n-1=3^a\cdot 5^b\cdot 7^c$
Since$$3||(2^2-1),$$$$5||(2^4-1),$$$$7||(2^3-1)$$Then through LTE we can know that :
If $2|n,a=v_3(2^n-1)=v_3(\frac n2)+1\leq \log_3\frac n2+1.$
If $4|n,b=v_5(2^n-1)=v_5(\frac n4)+1\leq \log_5\frac n4+1.$
If $3|n,c=v_7(2^n-1)=v_7(\frac n3)+1\leq \log_7\frac n3+1.$
Hence, $$2^n-1=3^a\cdot 5^b\cdot 7^c\leq 3^{\log_3\frac n2+1}\cdot 5^{\log_5\frac n4+1}\cdot 7^{\log_7\frac n3+1}=\frac{85n^3}{24}$$(If $a, b$ or $c=0$, the inequality above is still ture evidently.)
Since $n\in \mathbb{N}^+,$ the inequality above requires $n\leq12$.
We can verify that the original statement is TRUE only when $n=1,2,3,4,6$ .$\square$
This post has been edited 1 time. Last edited by kub-inst, Jun 5, 2024, 6:03 AM
Reason: mistyped
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Scilyse
387 posts
#20
Y by
Zsigmondeez nuts
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Eka01
204 posts
#21
Y by
By Zsigmondy, every $n$ gives a new prime factor apart from a small few (like greater than 10 or smth, dont have the energy to recall the edge cases). Then just check the few cases by hand.
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alexanderhamilton124
400 posts
#22
Y by
7 | 2^3 - 1 ==> we done by zsigmondy
This post has been edited 1 time. Last edited by alexanderhamilton124, Jan 9, 2025, 3:34 AM
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RedFireTruck
4243 posts
#24
Y by
n=1 gives 1
n=2 gives 3
n=3 gives 7
n=4 gives 15=3*5
n=6 gives 63=3*3*7

by zsigmondy theorem, every other 2^n-1 must have a prime factor other than 3,5,7 so this is all
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Ihatecombin
69 posts
#25
Y by
L.T.E also works here, even with brain dead level bounding one only needs to check until $n=14$.
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Giant_PT
50 posts
#26
Y by
Almost trivial by Zsigmondy :(
We can see that the smallest $n$ for which $3|2^{n}-1$, $5|2^{n}-1$, and $7|2^{n}-1$ are $2$, $4$ and $3$ respectively. Also, we have the exceptional case when $n=6$, so by just checking small values of $n$, we see that $n=1,2,3,4,6$ are the only solutions.
This post has been edited 2 times. Last edited by Giant_PT, Apr 22, 2025, 3:41 PM
Reason: Typos
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Assassino9931
1378 posts
#27
Y by
Here is a solution with very elementary methods only.

Direct verification shows that $n=1,2,3,4,6$ work and $n=5,7$ do not. We prove that no $n\geq 8$ works.

Note that $2^n - 1$ is divisible by $3$ if and only if $n$ is even, divisible by $5$ if and only if $n$ is a multiple of $4$, and divisible by $7$ if and only if $n$ is a multiple of $3$. Suppose $n$ is odd - then $2^n - 1 = 7^t$, which is impossible for $n\geq 4$ by mod $16$.

Hence $n$ is even. If $2^n - 1$ is divisible by all of $3,5,7$, then $n$ is divisible by $12$, but then $13$ divides $2^{n} - 1$ by Fermat, contradiction. Hence $2^n - 1 = 3^a, 5^a, 3^a5^b$ or $3^a7^b$. The first two are impossible by mod $8$ (the second one even by mod $4$). For $2^n - 1 = 3^a5^b$, $a,b > 0$, we must have $n=4t, t \geq 2$ and then $(2^t - 1)(2^t + 1)(2^{2t} + 1) = 3^a5^b$ where the factors on the left are pairwise coprime - hence at least one of them equals $1$, impossible for $t\geq 2$. Finally, for $2^n - 1 = 3^a7^b$ we have $n=6t$ where $t\geq 3$ is odd (as $12$ does not divide $n$ by above), and then $(2^t - 1)(2^{t} + 1)(2^{2t} + 2^t + 1)(2^{2t} - 2^t + 1) = 3^a7^b$, with the first three factors being pairwise coprime (only the second one is divisible by $3$ for odd $t$), hence one of them equals $1$, impossible for $t\geq 3$.
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