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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
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1:1 Physics Tutors
DinoDragon186   3
N 2 minutes ago by talhee
I am looking for 1:1 physics tutor.
I am a beginner in physics and am in 9th grade.
I want to make it to IPhO in the coming years.
3 replies
DinoDragon186
Dec 10, 2024
talhee
2 minutes ago
Looking for Physics or USAPhO Tutor
physicsplease   4
N 8 minutes ago by talhee
Hii I am looking for a USAPhO tutor for next year's season. I think I have tried literally everything possible to improve but I feel like I just hit a massive roadblock right now.

It would be ideal if I can find someone who have a lot of experience with physics olympiads. My goal is medal/gold in usapho next year, and I am very determined & willing to put in a lot of hours, especially more so in the summer. Please recommend anyone or dm in aops, thank you.

Have qualified usapho before (last year), took both physics c and sufficient higher math.
4 replies
physicsplease
Apr 11, 2025
talhee
8 minutes ago
9 ARML Location
deduck   35
N 3 hours ago by deduck
UNR -> Nevada
St Anselm -> New Hampshire
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Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
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deduck
Yesterday at 4:19 PM
deduck
3 hours ago
HCSSiM results
SurvivingInEnglish   57
N 3 hours ago by lpieleanu
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
57 replies
SurvivingInEnglish
Apr 5, 2024
lpieleanu
3 hours ago
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Sums of pairs in a sequence
tenniskidperson3   56
N Apr 6, 2025 by Marcus_Zhang
Source: USAJMO 2010, Problem 2
Let $n > 1$ be an integer. Find, with proof, all sequences $x_1 , x_2 , \ldots , x_{n-1}$ of positive integers with the following three properties:
(a). $x_1 < x_2 < \cdots < x_{n-1}$ ;
(b). $x_i + x_{n-i} = 2n$ for all $i = 1, 2, \ldots , n - 1$;
(c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i + x_j = x_k$.
56 replies
tenniskidperson3
Apr 29, 2010
Marcus_Zhang
Apr 6, 2025
Sums of pairs in a sequence
G H J
Source: USAJMO 2010, Problem 2
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kamatadu
480 posts
#44 • 1 Y
Y by HoripodoKrishno
I claim that the only sequence is $x_i=2i$. It is easy to see that such a sequence suffices the condition. Now we prove that this is the only one.

Firstly, for some $n$, we prove that $x_i=i\cdot x_1$. We proceed using induction. The base case $x_1=1\cdot x_1=x_1$ is clearly true. Now assume the induction hypothesis for some $k-1<n-1$, and we prove it for $k$.

Consider the following sequence.\[x_1+x_{n-k},x_2+x_{n-k},\ldots,x_{k-1}+x_{n-k}.\]Note that this sequence has $k-1$ terms in it. Firstly, we have that $x_k+x_{n-k}=2n$ and the fact that $x_i<x_k$ for all $1\le i\le k-1$. So this gives us that $x_i+x_{n-k}<x_k+x_{n-k}=2n$ which means that we will get some $x_j$ for each $i$ such that $x_j=x_i+x_{n-k}$. Now clearly from the increasing condition of the sequence, we get that $j\ge n-k+1$. So each $x_j$ belongs to the following sequence.\[x_{n-(k-1)},x_{n-(k)},\ldots,x_{n-1}.\]So note that this sequence also has $k-1$ terms. So this forces that the equality occurs in each of the values of $x_j$ from both the sequences. So this gives that $x_1+x_{n-k}=x_{n-(k-1)}$, which further gives that $x_1+(2n - x_{n-(n-k)})=2n - x_{n-((n-k+1)}\implies x_1 - x_k = -x_{k-1}\implies x_1+x_{k-1}=x_k$. Now using our induction hypothesis, we get that $x_{k-1}=(k-1)x_1$ which finally gives $x_k=k\cdot x_1$. Now then, we finally get that $x_{n-1}=(n-1)x_1$. Putting this into $x_1+x_{n-1}=2n$, we get that $x_1=2$ and we are done.
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trk08
614 posts
#45
Y by
We claim the only possible sequence is if $x_i=2i$. This sequence clearly works, and we now prove it is the only one.//

First of all, consider $x_1+x_{n-2}$. As $x_1>0$, it has to be more than $x_{n-2}$, which implies that it must be $x_{n-1}$. As each of the following sums are in decreasing order, from top to bottom:
\[x_1+x_{n-2}\]\[x_1+x_{n-3}\]\[\dots\]\[x_1+x_1,\]and there are $n-2$ of them, we can say that:
\[x_1+x_{n-2}=x_{n-1}\]\[x_1+x_{n-3}=x_{n-2}\]\[\dots\]\[x_1+x_1=x_2.\]

Therefore, it is easy to see that $x_{i}=ix_1$. Therefore, $nx_1=2n$, or $x_1=2$, as desired.
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Shreyasharma
682 posts
#46
Y by
Note that we are given,
  • $x_1 < x_2 < \dots < x_{n-1}$
  • $x_1 + x_{n-1} = x_2 + x_{n-2} = x_3 + x_{n-3} = \dots = 2n$
From the final condition we find $x_1 + x_2$, $x_1 + x_3$, $\dots$, $x_1+x_{n-2}$ all belong to $(x_i)$. Clearly $x_1 + x_{n-2} = x_{n-1}$ or that $$x_{n-2} = x_{n-1} - x_1$$Then $x_1 + x_{n-3} = x_{n-2}$, which in turn gives, $$x_{n-3} = x_{n-1} - 2x_1$$and so on. Inducting downwards we find $$x_{n-1-k} = x_{n-1} - kx_1$$However then $x_1 = x_{n-1} - (n-2)x_1$ which gives $nx_1 = 2n$, or $x_1 = 2$. Now from our induction we find the sequence of $(x_i)$ are simply the even integers from $2$ to $2n - 2$.
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joshualiu315
2534 posts
#47
Y by
Consider the sequence $x_1+x_1, x_1+x_2, \dots, x_1+x_{n-1} = 2n$. Each of the $n-2$ terms before $2n$ are obviously less than $2n$, and they must be part of the sequence $x_1, x_2, \dots, x_{n-1}$. This means we must necessarily have

\begin{align*}
    x_1+x_1 &= x_2, \\
    x_1+x_2 &= x_3, \\
    \dots &\phantom{=} \\
    x_1+x_{n-2} &= x_{n-1}.
\end{align*}
Hence, $x_k = kx_1$, and plugging this in gives

\[x_1+(n-1)x_1=2n \implies x_1=2.\]
Thus, $x_n = \boxed{2n}$, which is easily checked to be true.
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dolphinday
1324 posts
#48
Y by
We find that
\[x_1 < x_1 + x_1 < x_1 + x_2 \dots x_1 + x_{n-2}\]are all in $x_i$, due to $(c)$, and since there are $n-1$ terms, this sequence must be equivalent to
\[x_1, x_2, \dots, x_{n-1}\]So, we can rewrite each term as $ix_1$.
$\newline$
\[x_1 + x_{n-1} = 2n\]can be rewritten as
\[x_1 + (n-1)x_1 = 2n \implies nx_1 = 2n \implies x_1 = 2\]Due to the fact that each term is equal to $ix_1$, this sequence is just $x_i = 2i$.
This post has been edited 1 time. Last edited by dolphinday, Jan 6, 2024, 11:58 AM
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shendrew7
795 posts
#49
Y by
Due to the third condition, we know the sequence $x_2, x_3, \ldots, x_{n-1}, 2n$ must correspond with the terms of
\[x_1+x_1 < x_1+x_2 < \ldots < x_1+x_{n-2} < x_1+x_{n-1} = 2n.\]
Thus we know $x_k = k \cdot x_1$ for $1 \leq k \leq n-1$, from which the second condition tells us our only solution is $\boxed{x_i = 2i}$. $\blacksquare$
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peppapig_
281 posts
#50
Y by
I goofed last time;

I claim that the only possible sequence is $2$, $4$, $6$, $\dots$, $2n-2$. Note that by problem conditions, we have that
\[x_1+x_{n-1}=2n,\]but we also have that
\[x_1+x_{n-2}=x_j,\]for some integer $j$. However, since $x_1<x_2<\dots<x_{n-1}$, we must have that $x_{n-2}=x_{n-1}-x_1$. Similarly, we get that
\[x_k=x_1+x_{n-3}<x_1+x_{n-2}=x_{n-1},\]meaning that $k=n-2$. Continuing this, we get that our sequence must be $x_1$, $2x_1$, $\dots$, $(n-1)x_1$, and since $x_1+x_{n-1}=2n$, we also get that $x_1$ must be equal to $2$. Therefore the only possible sequence is $2$, $4$, $6$, $\dots$, $2n-2$, which indeed works, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Feb 24, 2024, 3:33 AM
Reason: Missed a $
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two_steps
102 posts
#51
Y by
is this correct? why is my solution different from everyone else lol

For any two terms $x_i < x_j$, notice that the following terms are also in the sequence
\[x_i, x_j \rightarrow 2n-x_j \rightarrow 2n-x_j+x_i \rightarrow x_j-x_i\]This is essentially a stronger version of the converse of condition 3. From here, note that we may prove that each element is a multiple of $x_1$ by induction. The only possible sequences are now $1,2,\dots, n-1$ and $2,4,\dots,2(n-1)$. The first sequence doesn't work, while the second one does.
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ItsBesi
146 posts
#52
Y by
My solution is different from others so did I fakesolve?

We separate the problem into $2$ cases which both are analogous.

Case 1. $n-\text{is even}$
Click to reveal hidden text

Case 2. $n-\text{is odd}$
Click to reveal hidden text

Since in both cases we have: $\{x_1,x_{k_1}, \dots , x_{k_{n-2}} \}=\{x_1,x_2, \dots, x_{n-1} \}$ we get:

$x_1=x_1 , x_2=x_{k_1} , x_3=x_{k-2} ,  \dots , x_{n-1}=x_{k_{n-2}}$
So
$x_1+x_1=x_{k_1}=x_2 \implies x_1+x_1=x_2 \implies x_2=2 \cdot x_1$
$x_1+x_2=x_{k_2}=x_3 \implies x_1+x_2=x_3 \implies x_3=x_1+x_2=x_1 + 2 \cdot x_1=3 \cdot x_1 \implies x_3=3 \cdot x_1$
$x_2+x_2=x_{k_3}=x_4 \implies x_2+x_2=x_4 \implies x_4=2 \cdot x_2=2 \cdot 2 \cdot x_1= 4 \cdot x_1 \implies x_4=4 \cdot x_1$
$x_2+x_3=x_{k_4}=x_5 \implies x_2+x_3=x_5 \implies x_5=x_2+x_3=2 \cdot x_1 + 3 \cdot x_1 =5 \cdot x_1 \implies x_5= 5 \cdot x_1$
$x_3+x_3=x_{k_5}=x_6 \implies x_3+x_3=x_6\implies x_6=x_3+x_3=3 \cdot x_1 + 3 \cdot x_1=6 \cdot x_1 \implies x_6=6 \cdot x_1$
$\dots$
$x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=x_{k_{n-3}}=x_{n-2} \implies x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=x_{n-2} \implies x_{n-2}= x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=2 \cdot  x_{\frac{n}{2}-1} = 2 \cdot (\frac{n}{2}-1) \cdot x_1=(n-2) \cdot x_1 \implies x_{n-2}=(n-2) \cdot x_1$

Now again by $(b) \implies$

$x_2+x_{n-2}=2n \implies 2 \cdot x_1 + (n-2) \cdot x_1 =2n \implies n \cdot x_1=2n \implies x_1=2
So x_1=2, x_2=2 \cdot x_1=2 \cdot 2=4 , x_3=3 \cdot x_1= 3 \cdot 2=6 , \dots x_{n-1}=(n-1) \cdot x_1=(n-1) \cdot 2=2n-2$

Hence $(x_1,x_2, \dots , x_{n-1}=(2,4, \dots ,2n-2)$
This post has been edited 1 time. Last edited by ItsBesi, Sep 26, 2024, 1:12 PM
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Maximilian113
575 posts
#53
Y by
Observe that property (b) is equivalent to $$i+j < n \iff x_i+x_j < 2n.$$Therefore, $$x_1 < x_1+x_1 < x_1+x_2 < \cdots < x_1+x_{n-2} < 2n.$$However, each of these sums are some $x_k$ but since $$x_2<x_3<\cdots < x_{n-1} < 2n$$it follows that $x_1+x_k=x_{k+1}$ for each $k=1, 2, \dots, n-2.$ Thus $x_2=2x_1, x_3=3x_1,$ etc. This yields $2n=x_1+x_{n-1}=x_1+(n-1)x_1 \implies x_1=2,$ and it follows that the sequence is $$2, 4, 6, \cdots, 2n-2.$$It is easy to show that this works.
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eg4334
637 posts
#54
Y by
Goofy problem.

The conditions imply that the sums, in increasing order, $\{ x_1+x_1, x_1+x_2, \dots x_1+x_{n-1} \}$ must biject to $x_2, x_3, \dots x_{n-1}$. Therefore, $x_1+x_1=x_2$, $x_1+x_2=x_3$, and likewise. Then its not hard to see that $x_i = i x_1$ and from the second condition we obtain $x_1=2$. Therefore the only sequence is the increasing one of evens.
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Ihatecombin
60 posts
#55
Y by
Notice that \(a_1,a_1 + a_1, a_1 + a_2, \dots. a_1 +a_{n-2}\) are all different,
by an easy induction we obtain the only sequence that works is \(a_i = 2i\).
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bjump
1023 posts
#56 • 2 Y
Y by KenWuMath, imagien_bad
Solution from over a year ago ...

The answer is $x_{k}= 2n$
Note that
$$x_1 + x_{n-1}=2n$$$$x_{1}+ x_{n-2} = x_{n-1}$$$$\vdots$$$$x_{1}+x_{1}=x_{2}$$I claim that $x_{n} = nx_1$ we will prove this via induction with a trivial base case of $n=1$. Assume that this is true for $1 \le n \le k$ then $x_{k+1} = x_{k}+ x_1=kx_1+x_1= (k+1)x_1$. Now by condition (b) $x_i+x_{n-i} = 2n \iff (n-i+i)x_1=2n \iff x_1=2$. Therefore $x_k = k x_1 = 2k$.
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de-Kirschbaum
198 posts
#57
Y by
Note that $x_{n-2}<x_1+x_{n-2}<x_2+x_{n-2}=2n$ so by property 3 we must have $x_{n-2}+x_1=x_{n-1}$. By a similar reasoning this works for all $x_i$ so we have that $x_1, x_2,\ldots, x_{n-1}$ is an arithmetic sequence with common difference $x_1$, and by property 1 we must have $nx_1=2n \implies x_1=2$ so the only sequence that works is $x_i=2i$.
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Marcus_Zhang
980 posts
#58
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Storage
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