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jlacosta   0
Yesterday 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.

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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
Austrian Regional MO 2025 P4
BR1F1SZ   2
N a few seconds ago by NumberzAndStuff
Source: Austrian Regional MO
Let $z$ be a positive integer that is not divisible by $8$. Furthermore, let $n \geqslant 2$ be a positive integer. Prove that none of the numbers of the form $z^n + z + 1$ is a square number.

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
a few seconds ago
Austrian Regional MO 2025 P3
BR1F1SZ   1
N 2 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
There are $6$ different bus lines in a city, each stopping at exactly $5$ stations and running in both directions. Nevertheless, for every two different stations there is always a bus line connecting these two stations. Determine the maximum number of stations in this city.

(Karl Czakler)
1 reply
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
2 minutes ago
Austrian Regional MO 2025 P2
BR1F1SZ   2
N 7 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
Let $\triangle{ABC}$ be an isosceles triangle with $AC = BC$ and circumcircle $\omega$. The line through $B$ perpendicular to $BC$ is denoted by $\ell$. Furthermore, let $M$ be any point on $\ell$. The circle $\gamma$ with center $M$ and radius $BM$ intersects $AB$ once more at point $P$ and the circumcircle $\omega$ once more at point $Q$. Prove that the points $P,Q$ and $C$ lie on a straight line.

(Karl Czakler)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
7 minutes ago
Austrian Regional MO 2025 P1
BR1F1SZ   2
N 8 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
Let $n \geqslant 3$ be a positive integer. Furthermore, let $x_1, x_2,\ldots, x_n \in [0, 2]$ be real numbers subject to $x_1 + x_2 +\cdots + x_n = 5$. Prove the inequality$$x_1^2 + x_2^2 + \cdots + x_n^2 \leqslant 9.$$When does equality hold?

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
8 minutes ago
Putnam 1954 A1
sqrtX   2
N 43 minutes ago by centslordm
Source: Putnam 1954
Let $n$ be an odd integer greater than $1.$ Let $A$ be an $n\times n$ symmetric matrix such that each row and column consists of some permutation of the integers $1,2, \ldots, n.$ Show that each of the integers $1,2, \ldots, n$ must appear in the main diagonal of $A$.
2 replies
sqrtX
Jul 17, 2022
centslordm
43 minutes ago
Putnam 1953 B1
sqrtX   7
N 44 minutes ago by centslordm
Source: Putnam 1953
Is the infinite series
$$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$$convergent?
7 replies
sqrtX
Jul 16, 2022
centslordm
44 minutes ago
1953 Putnam A2
Taco12   4
N an hour ago by centslordm
Source: 1953 Putnam A2
The complete graph with 6 points and 15 edges has each edge colored red or blue. Show that we can find 3 points such that the 3 edges joining them are the same color.
4 replies
Taco12
Aug 20, 2021
centslordm
an hour ago
Putnam 1952 B4
sqrtX   1
N an hour ago by centslordm
Source: Putnam 1952
A homogeneous solid body is made by joining a base of a circular cylinder of height $h$ and radius $r,$ and the base of a hemisphere of radius $r.$ This body is placed with the hemispherical end on a horizontal table, with the axis of the cylinder in a vertical position, and then slightly oscillated. It is intuitively evident that if $r$ is large as compared to $h$, the equilibrium will be stable; but if $r$ is small compared to $h$, the equilibrium will be unstable. What is the critical value of the ratio $r\slash h$ which enables the body to rest in neutral equilibrium in any position?
1 reply
sqrtX
Jul 7, 2022
centslordm
an hour ago
Putnam 1952 B3
centslordm   2
N an hour ago by centslordm
Develop necessary and sufficient conditions that the equation \[ \begin{vmatrix} 0 & a_1 - x & a_2 - x \\ -a_1 - x & 0 & a_3 - x \\ -a_2 - x & -a_3 - x & 0\end{vmatrix} = 0 \qquad (a_i \neq 0) \]shall have a multiple root.
2 replies
centslordm
May 30, 2022
centslordm
an hour ago
Putnam 1952 A6
centslordm   1
N an hour ago by centslordm
A man has a rectangular block of wood $m$ by $n$ by $r$ inches ($m, n,$ and $r$ are integers). He paints the entire surface of the block, cuts the block into inch cubes, and notices that exactly half the cubes are completely unpainted. Prove that the number of essentially different blocks with this property is finite. (Do not attempt to enumerate them.)
1 reply
centslordm
May 29, 2022
centslordm
an hour ago
Putnam 1952 A4
centslordm   2
N an hour ago by centslordm
The flag of the United Nations consists of a polar map of the world, with the North Pole as its center, extending to approximately $45^\circ$ South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude $30^\circ$ S.In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?
2 replies
centslordm
May 29, 2022
centslordm
an hour ago
Putnam 1958 November A7
sqrtX   1
N 4 hours ago by centslordm
Source: Putnam 1958 November
Let $a$ and $b$ be relatively prime positive integers, $b$ even. For each positive integer $q$, let $p=p(q)$ be chosen so that
$$ \left| \frac{p}{q} - \frac{a}{b}  \right|$$is a minimum. Prove that
$$ \lim_{n \to \infty} \sum_{q=1 }^{n} \frac{ q\left| \frac{p}{q} - \frac{a}{b}  \right|}{n} = \frac{1}{4}.$$
1 reply
sqrtX
Jul 19, 2022
centslordm
4 hours ago
Putnam 1958 November B7
sqrtX   5
N 4 hours ago by centslordm
Source: Putnam 1958 November
Let $a_1 ,a_2 ,\ldots, a_n$ be a permutation of the integers $1,2,\ldots, n.$ Call $a_i$ a big integer if $a_i >a_j$ for all $i<j.$ Find the mean number of big integers over all permutations on the first $n$ postive integers.
5 replies
sqrtX
Jul 19, 2022
centslordm
4 hours ago
System of two matrices of the same rank
Assassino9931   3
N 5 hours ago by RobertRogo
Source: Vojtech Jarnik IMC 2025, Category II, P2
Let $A,B$ be two $n\times n$ complex matrices of the same rank, and let $k$ be a positive integer. Prove that $A^{k+1}B^k = A$ if and only if $B^{k+1}A^k = B$.
3 replies
Assassino9931
Today at 1:02 AM
RobertRogo
5 hours ago
IMO problem 1
iandrei   77
N Apr 23, 2025 by YaoAOPS
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
77 replies
iandrei
Jul 14, 2003
YaoAOPS
Apr 23, 2025
IMO problem 1
G H J
Source: IMO ShortList 2003, combinatorics problem 1
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Math4Life7
1703 posts
#69
Y by
We can see that for $a_x, a_y \in A$. We want no pairs such that $a_x - a_y = t_i - t_j$.

We choose our $t_i$ greedily and by induction. Firstly we choose $t_1 =1$.

For our induction step we let $t_{n+1}$ be the first value such that $t_{n+1} - t_j \neq a_x - a_y$ for all $j \in [1, n]$. We can see that there are $\binom{101}{2}$ values of $a_x - a_y$, and there are $n$ values of $j$. Thus the number of values that cannot be obtained by $t_{100}$ is bounded above by $\binom{101}{2} \cdot 100 \leq 10^6$. $\blacksquare$
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ihatemath123
3446 posts
#70
Y by
The "disjoint" condition is satisfied iff $a_w + t_x \neq a_y + t_z$ for all valid $(w,x,y,z)$; in other words, $|a_w - a_y| \neq |t_x - t_z|$ for all valid $w, x, y$ and $z$. There are up to $\binom{101}{2} = 5050$ distinct absolute differences between two elements of $A$.

Consider just arithmetic sequences $t_1, t_2, \dots, t_{100}$ with common difference $d$, where $1 \leq d \leq 10101$. The possible values of $|t_x - t_z|$ are $\{ d, 2d, \dots, 99d \}$; as $d$ varies from $1$ to $10101$, there must be some (in fact, at least $5051$) values of $d$ for which none of $\{ d, 2d, \dots, 99d \}$ are the absolute difference between any two elements of $A$, so we just pick $d$ to be any of those values.
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mannshah1211
651 posts
#71
Y by
Let $k$ be the largest size of such a set $t$. Let $t_1, t_2, \cdots, t_k$ be the numbers, and let $A_1, A_2, \cdots, A_{101}$ be the numbers. Let $n$ be the number of numbers of the form $x - y + t_j$, where $x, y  \in A$. Then, we must have the inequality $10^6 - k \le n,$ since if $10^6 - k > n,$ we have at least one more valid choice to add to $t$. But, in turn, $n \le 101 \cdot 100 \cdot k$, which gives the desired inequality.

EDIT: This is incorrect. I'll edit in the correct solution sometime soon.
EDIT 2: Done.
This post has been edited 3 times. Last edited by mannshah1211, Jan 17, 2024, 9:53 AM
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eibc
600 posts
#72
Y by
Suppose that we pick $n$ numbers $t_1, t_2, \ldots, t_n$ satisfying the problem conditions for maximal $n$. Then for $x \in S$ and $x \neq t_k$ for some $k$, we must have $x - t_{\ell} = a_i - a_j$ or $x = t_{\ell} + a_i - a_j$ for some integers $1 \le \ell \le n$ and $1 \le i \neq j \le 100$. For a fixed $t_\ell$, there are $101 \cdot 100 = 10100$ ways to choose $(a_i, a_j)$, which is enough to imply that $10100n \ge 10^6 - n$, or $n \ge 100$.
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RGB
19 posts
#73
Y by
We can prove this by contradiction. Suppose there is a $101$-element subset $A$ of $S = \{1,2,\ldots,1000000\}$ such that, for any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ from $S$, there exist indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $A_i$ and $A_j$ have at least one common element.

Let $B_1, B_2, \ldots, B_{100}$ be the sets $A_1, A_2, \ldots, A_{100}$ respectively. For each $j$, the set $B_j$ can be expressed as $B_j = A + t_j = \{x + t_j \mid x \in A\}$.

Now, consider the union of all $B_j$'s for $j = 1, 2, \ldots, 100$:

\[ B = B_1 \cup B_2 \cup \ldots \cup B_{100} \]
Since $B_j = A + t_j$, the union $B$ is essentially the set of all elements in $S$ that can be obtained by adding an element from $A$ to any of the $100$ chosen $t_j$'s.

Now, by our assumption, there must exist two indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $B_i \cap B_j \neq \emptyset$. This implies that there exists an element $x$ in $A$ such that $x + t_i = x + t_j$, which means $t_i = t_j$.

However, this contradicts our assumption that any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ must yield distinct sets $B_j$. Therefore, our assumption must be false, and there must be a $101$-element subset $A$ with the desired property.
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dolphinday
1324 posts
#74
Y by
Keep on choosing $t_i$ until we get stuck(we can't pick any more $t$) for some $i = k$. This implies that for all $t \in S$, it has already been picked or that for some $a$, $b$, $t_i \in S$ we have $t = b + t_i - a$. So then there are at most $k \cdot 10100$($101$ choices for $b$ and $100$ for $a$) values for $t$, so $k \geq 100$ as desired.
This post has been edited 1 time. Last edited by dolphinday, Feb 25, 2024, 7:21 PM
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bjump
1013 posts
#75 • 1 Y
Y by megarnie
badly worded solution
Choose $t_i$'s greedily then for all $a$, $b$ in $A$, and $1 \le j \le 100$ we cannot pick $t_j+a-b$, or $t_i-b+a$ in the list of $t$'s in the future. So for every new $t$ added we eliminate at most $101 \cdot 100 = 10100$ more $t$'s. So we can pick at least $\left \lceil \frac{10^6}{10100} \right \rceil = 100$ different $t$'s before we could be forced to stop, finishing the problem.
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RedFireTruck
4221 posts
#76
Y by
we basically just dont want any absolute difference in $A$ to also be an absolute difference in $t$

there are at most $\binom{101}{2}=5050$ distinct absolute differences in $A$

this means that every time we pick a $t_i$, at most $1+5050\cdot2=10101$ elements get eliminated from $S$ as possibilities for other $t_i$

if we pick $t_1=1$, then only at most $5051$ elements get eliminated from $S$ the first time

since $5051+98\cdot 10101=989898+5051<995000$, we can pick $t_{100}$, as desired.
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ezpotd
1262 posts
#77
Y by
Observe the condition is equivalent to the difference between no two $t_i$ being the difference between two $a_i$. Choose $t_i$, for each $t_i$ we automatically eliminate out of candidacy for future $t_i$ all the numbers that are a different of two $A_i$ from $t_i$, which is at most $2 \cdot \binom{101}{2}$, where the $2$ is for positve/negative, and we also eliminate the chosen number itself, for a total for $10101$ eliminations per chosen $t_i$. Now any number not eliminated can be added to $t_i$ and the condition will still be satisfied, so add the first $99$ numbers and see that we have eliminated $999999$ numbers, and there is one number remaining and we see that we can just add this and done.
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joshualiu315
2533 posts
#78
Y by
Suppose we have a working set: $A = \{a_1, a_2, \dots, a_{101} \}$. Note that $A_i$ and $A_j$ are disjoint if and only if

\[t_i-t_j \notin \{a_p-a_q\vert 1\le p \ne q \le 101\}.\]
In order to get such a set, we will proceed by a greedy algorithm. Assume that we have already picked $n$ elements in $T = \{t_1, \dots, t_n\}$ such that $T$ is as large as possible. That means every $i \in \{1,2,\dots,10^6\}$ is either in $T$ or satisfies

\[i = t_i+b-a,\]
for some $t_i \in T$ and $a,b \in A$.

There are at most $|T| \cdot |A| \cdot (|A|-1) = 10100n$ values of $i$. Thus,

\[n+10100n \ge 10^6 \implies n>99,\]
as desired. $\blacksquare$
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Saucepan_man02
1328 posts
#79
Y by
Here we go:
Consider the set $T$ to have $t_1, \cdots, t_n$ currently. Let $A = \{ a_1, a_2, \cdots, a_{101} \}$. Let set $D$ denote the pair-wise differences of elements of set $A$.
Then, if we cant add another element $t \in S$ to $S$, then we must have: $t = t_k + d$ for some $d \in D$. Note that: $|D| \le 101*100$. Thus: $$101*100*n > 10^6 \implies n \ge 100$$and we are done.
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Maximilian113
574 posts
#80
Y by
Let the elements of $A$ be $a_1, a_2, \cdots a_{101},$ and let the elements of $T$ be $t_1, t_2, \cdots, t_n.$ Since each $x+t_j, x\in A$ is distinct, then clearly for all $p, q, r, s$ $$|t_p-t_q| \neq |a_r-a_s|.$$Set $t_1=1.$ Clearly since there are at most $\binom{101}{2}=5050$ pairwise differences among the elements of $A,$ every time we add a number to the set $T$ we reduce the possible number of potential numbers to add to $T$ by at most $2 \cdot 5050 + 1 = 10101.$ (we account for both sides, and the number itself) Therefore, after choosing $99$ numbers, there are at least $$10^6 - 10101\cdot 99=1$$possible number to add to the set $T,$ so thus it is possible to have $100$ elements in $T.$ QED
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megahertz13
3183 posts
#81
Y by
First choose a random $t_1\in S$.

We will prove that if we have already chosen distinct $t_1,t_2,\dots,t_k$ satisfying the condition, then we can choose a $t_{k+1}$ such that the disjoint condition is satisfied. This will finish the problem.

If there are any intersections, there exists a $1\le j\le k$ and $x,y\in A$ satisfying $$t_j+x=t_{k+1}+y\implies t_{k+1}=t_j+x-y.$$However, since $t_j\ne t_{k+1}$, we have $x\ne y$. Then, there are $$k\cdot 101\cdot 100=10100k$$ways to choose the right-hand side of the equation $t_{k+1}=t_j+x-y$. Since $t_{k+1}$ is not equal to any of $t_1, t_2,\dots,t_k$, there are at most $10101k$ numbers that $t_{k+1}$ cannot be. However, $$10101k\le 999999<|S|,$$so there always exists a $t_{k+1}$.
This post has been edited 1 time. Last edited by megahertz13, Nov 26, 2024, 5:05 PM
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ihategeo_1969
225 posts
#82
Y by
Say $t_1$, $\dots$, $t_k$ are max such subsets.

So adding $A_{k+1}$ has an element common with one of $A_1$, $A_2$, $\dots$, $A_k$. Hence for any $T \in S/\{t_1\dots,t_k\}$, we have \[a_\ell+T=a_j+t_i\]Fix RHS and see that $a_\ell$ takes $101-1$ values and so \[101k \ge \frac{10^6-k}{100} \implies k>99 \iff k \ge 100\]And done.
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YaoAOPS
1535 posts
#83
Y by
Let $D = A - A$ be the elements which can be expressed as the difference of two elements in the list. Then $|S| \le 101 \cdot 100 + 1 = 10101$ since $0 \in D$ is over counted $100$ times. We now select each $t_i$ iteratively. Note that if $t_i$ is selected, then any subsequent $t_j$ selected can't be in $t_i + D$ for the sets to be pairwise disjoint, and furthermore this is sufficient. As such, we may select $t_i$ from $S \setminus \bigcup_{j=1}^{i-1} (t_j + D)$ while preserving the condition. For each $i$, we have that
\begin{align*}
	\left|S \setminus \bigcup_{j=1}^{i-1} (t_j + D)\right| &\ge
	1000000 - \sum_{j=1}^{99} \left|(t_j + D)\right|  \\
    &\ge 1000000 - 99 \cdot 10101 = 1.
\end{align*}so we may choose $t_{100}$ as well to finish.
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