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Difference between revisions of "2007 USAMO Problems/Problem 1"
(what I wrote (am I right?))
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== Solution ==
 
== Solution ==
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Suppose we create a parallel integer sequence <math>\displaystyle b_1, b_2, \ldots</math> such that for every <math>\displaystyle k \ge 1</math>, we have that <math>b_k = \frac{\displaystyle \sum_{i=1}^{k}}{k}</math>. Consider what happens when <math>b_k \le k</math>. For <math>\displaystyle k + 1</math>, we have that <math>b_{k+1} = \frac{\displaystyle \sum_{i=1}^{k+1}}{k+1} = \frac{\displaystyle \sum_{i=1}^{k} + a_{k+1}}{k+1} = \frac{b_k \times k + a_{k+1}}{k+1}</math>. <math>\displaystyle b_k</math> is a permissible value of <math> a_{k+1} \displaystyle </math> since <math>b_k \le k \le (k+1)-1</math>: if we substitute <math>\displaystyle b_k</math> for <math> a_{k+1} \displaystyle </math>, we get that <math>b_{k+1} = \frac{b_k (k+1)}{k+1} = b_k</math>. <math>\displaystyle b_k</math> is the unique value for <math> a_{k+1} \displaystyle </math>. We can repeat this argument for <math>\displaystyle b_k = b_{k+1} = b_{k+2} \ldots</math>. As we substituted the <math>\displaystyle b_k</math>s for the <math>\displaystyle a_k</math>s, the <math>\displaystyle a_k</math>s also become constant.
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Now we must show that <math>\displaystyle b_k</math> eventually <math>\le k</math>. Suppose that <math>\displaystyle b_k</math> always <math>\displaystyle > k</math>. By definition, <math> \frac{\displaystyle \sum_{i=1}^{k}}{k} = b_k > k</math>, so <math>\displaystyle \sum_{i=1}^{k} a_i > k^2</math>. We also have that each <math>a_i \le i-1</math> so that <math>k^2 < \displaystyle \sum_{i=1}^{k} \le n + 1 + 2 + \ldots + (k-1) = n + \frac{k^2 - k}2 </math>. So <math>k^2 < n +\frac{k^2 - k}2  \Longrightarrow \frac{k^2 + k}2 < n</math>. But <math>\displaystyle n</math> is constant while <math>\displaystyle k</math> is increasing, so eventually we will have a contradiction and <math>b_k \le k</math>. Therefore, the sequence of <math>\displaystyle a_i</math>s will become constant.
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{{USAMO newbox|year=2007|before=First question|num-a=2}}
 
{{USAMO newbox|year=2007|before=First question|num-a=2}}

Revision as of 17:46, 25 April 2007

Problem

Let $n$ be a positive integer. Define a sequence by setting $a_1 = n$ and, for each $k>1$, letting $a_k$ be the unique integer in the range $0 \le a_k \le k-1$ for which $a_1 + a_2 + \cdots + a_k$ is divisible by $k$. For instance, when $n=9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \ldots$. Prove that for any $n$ the sequence $a_1, a_2, a_3, \ldots$ eventually becomes constant.

Solution

Suppose we create a parallel integer sequence $\displaystyle b_1, b_2, \ldots$ such that for every $\displaystyle k \ge 1$, we have that $b_k = \frac{\displaystyle \sum_{i=1}^{k}}{k}$. Consider what happens when $b_k \le k$. For $\displaystyle k + 1$, we have that $b_{k+1} = \frac{\displaystyle \sum_{i=1}^{k+1}}{k+1} = \frac{\displaystyle \sum_{i=1}^{k} + a_{k+1}}{k+1} = \frac{b_k \times k + a_{k+1}}{k+1}$. $\displaystyle b_k$ is a permissible value of $a_{k+1} \displaystyle$ since $b_k \le k \le (k+1)-1$: if we substitute $\displaystyle b_k$ for $a_{k+1} \displaystyle$, we get that $b_{k+1} = \frac{b_k (k+1)}{k+1} = b_k$. $\displaystyle b_k$ is the unique value for $a_{k+1} \displaystyle$. We can repeat this argument for $\displaystyle b_k = b_{k+1} = b_{k+2} \ldots$. As we substituted the $\displaystyle b_k$s for the $\displaystyle a_k$s, the $\displaystyle a_k$s also become constant.

Now we must show that $\displaystyle b_k$ eventually $\le k$. Suppose that $\displaystyle b_k$ always $\displaystyle > k$. By definition, $\frac{\displaystyle \sum_{i=1}^{k}}{k} = b_k > k$, so $\displaystyle \sum_{i=1}^{k} a_i > k^2$. We also have that each $a_i \le i-1$ so that $k^2 < \displaystyle \sum_{i=1}^{k} \le n + 1 + 2 + \ldots + (k-1) = n + \frac{k^2 - k}2$. So $k^2 < n +\frac{k^2 - k}2 \Longrightarrow \frac{k^2 + k}2 < n$. But $\displaystyle n$ is constant while $\displaystyle k$ is increasing, so eventually we will have a contradiction and $b_k \le k$. Therefore, the sequence of $\displaystyle a_i$s will become constant.


2007 USAMO (Problems • Resources)
Preceded by
First question 		Followed by
Problem 2
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All USAMO Problems and Solutions
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