2010 USAJMO Problems/Problem 2

Problem

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:

  1. (a). $x_1 < x_2 < \cdots <x_{n-1}$;
  2. (b). $x_i +x_{n-i} = 2n$ for all $i=1,2,\ldots,n-1$;
  3. (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 $x_i+x_j = x_k$.

Solution

The sequence is $2, 4, 6, \ldots, 2n-2$.

Proof 1

We will prove that any sequence $x_1, \ldots, x_{n-1}$, that satisfies the given conditions, is an arithmetic progression with $x_1$ as both the first term and the increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$. Therefore $x_1 = 2$, and the sequence is just the even numbers from $2$ to $2n-2$. The sequence of successive even numbers clearly satisfies all three conditions, and we are done.

First a degenerate case. If $n = 2$, there is only one element $x_1$, and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$. Conditions (a) and (c) are vacuously true.

Otherwise, for $n > 2$, we will prove by induction on $m$ that the difference $x_{n-m} - x_{n-1-m} = x_1$ for all $m \in [1, n-2]$, which makes all the differences $x_{n-1} - x_{n-2} = \ldots = x_2 - x_1 = x_1$, i.e. the sequence is an arithmetic progression with $x_1$ as the first term and increment as promised.

So first the $m=1$ case. With $n > 2$, $x_{n-2}$ exists and is less than $x_{n-1}$ by condition (a). Now since by condition (b) $x_1 + x_{n-1} = 2n$, we conclude that $x_1 + x_{n-2} < 2n$, and therefore by condition (c) $x_1 + x_{n-2} = x_k$ for some $k$. Now, since $x_1 > 0$, $x_k > x_{n-2}$ and can only be $x_{n-1}$. So $x_1 + x_{n-2} = x_{n-1}$.

Now for the induction step on all values of $m$. Suppose we have shown that for all $i \le m$, $x_1 + x_{n-1-i} = x_{n-i}$. If $m = n-2$ we are done, otherwise $m < n-2$, and by condition (c) $x_1 + x_{n-2-m} = x_k$ for some $k$. This $x_k$ is larger than $x_{n-2-m}$, but smaller than $x_1 + x_{n-1-m} = x_{n-m}$ by the inductive hypothesis. It then follows that $x_1 + x_{n-2-m} = x_{n-1-m}$, the only element of the sequence between $x_{n-2-m}$ and $x_{n-m}$. This establishes the result for $i=m+1$.

So, by induction $x_1 + x_{n-1-m} = x_{n-m}$ for all $m \in [1, n-2]$, which completes the proof.

Proof 2

Let $S=\{x_1,x_2,...,x_{n-1}\}$. Notice that \[x_1<x_1+x_1<x_1+x_2<\dots <x_1+x_{n-2}<2n.\] Then by condition (c), we must have $x_1,x_1+x_1,...,x_1+x_{n-2}\in S$. This implies that $x_1=x_1,x_1+x_1=x_2,...,x_1+x_{n-2}=x_{n-1}$, or that $x_k=kx_1$. Then we have $x_1+x_{n-1}=n(x_1)=2n\rightarrow x_1=2$, and the rest is trivial.

Solution 2

The claim is that in this sequence, if there are $2$ elements $a,b$ where $a,b<n$, such that $\gcd(a,b)=1$, then the sequence contains every number less than $2n$.

Proof: Let $a$ and $b$ be the numbers less than $n$ such that $\gcd(a,b)=1$. We take this sequence modulo $n$. This means that if $x_i$ is an element in this sequence then $-x_i$ is as well. $a,b,2n-a,2n-b$ are all elements in the sequence. Clearly, one of $2n-a+b$ and $2n-b+a$ is less than $2n$, which means that $\pm (a-b)$ are in this sequence modulo $n$. Now we want to show every number is achievable. We have already established that $a$ and $b$ are relatively prime, so by euclidean algorithm, if we take the positive difference of $a'$ and $b'$ every time, we will get that $1$ is in our sequence. Then, we can simply add or subtract $1$ as many times from $a$ as desired to get every single number.

We have proved that there are no two numbers that can be relatively prime in our sequence, implying that no two consecutive numbers can be in this sequence. Because our sequence has $n-1$ terms, our sequence must be one of $2,4,6,..,2(n-1)$ or $1,3,5,...$, the latter obviously fails, so $2,4,6...2(n-1)$ is our only possible sequence.

See Also

2010 USAJMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAJMO Problems and Solutions

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