Difference between revisions of "2007 USAMO Problems/Problem 1"

(Solution 3)
(Solution 3)
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=== Solution 3 ===
 
=== Solution 3 ===
 
Let <math>a_1=n</math>.  Since <math>a_k\le k-1</math>, we have that <math>a_1+a_2+a_3+\hdots+a_n\le n+1+2+\hdots+n-1</math>.   
 
Let <math>a_1=n</math>.  Since <math>a_k\le k-1</math>, we have that <math>a_1+a_2+a_3+\hdots+a_n\le n+1+2+\hdots+n-1</math>.   
 +
 
Thus, <math>a_1+a_2+\hdots+a_n\le \frac{n(n+1)}{2}</math>.   
 
Thus, <math>a_1+a_2+\hdots+a_n\le \frac{n(n+1)}{2}</math>.   
 +
 
Since <math>a_1+a_2+\hdots+a_n=nk</math>, for some integer <math>k</math>, we can keep adding <math>k</math> to satisfy the conditions, provided that <math>k\le n</math> because <math>a_n+1\le n</math>.
 
Since <math>a_1+a_2+\hdots+a_n=nk</math>, for some integer <math>k</math>, we can keep adding <math>k</math> to satisfy the conditions, provided that <math>k\le n</math> because <math>a_n+1\le n</math>.
  

Revision as of 10:12, 1 April 2008

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

Solution 1

Define $s_k =\sum_{i=1}^{k} a_i$, and $b_k = \frac{s_k}{k}$. If $b_k \le k$, then for $k + 1$, $b_{k+1} = \frac{s_{k+1}}{k + 1} = \frac{s_k + a_{k+1}}{k+1} = \frac{b_k \cdot k + a_{k+1}}{k+1}$. Note that $b_k$ is a permissible value of $a_{k+1}$ since $b_k \le k = (k+1)-1$: if we substitute $b_k$ for $a_{k+1}$, we get $b_{k+1} = \frac{b_k (k+1)}{k+1} = b_k$, the unique value for $a_{k+1}$. So $b_k = b_{k+1} = b_{k+2} = \cdots$, from which if follows that the $a_k$s become constant.

Now we must show that eventually $b_k \le k$. Suppose that $b_k > k$ for all $k$. By definition, $\frac {s_k}{k} = b_k > k$, so $s_k > k^2$. Also, for $i>1$, each $a_i \le i-1$ so

$k^2 <  s_k \le n + 1 + 2 + \cdots + (k-1) = n + \frac{k^2 - k}2$
$k^2 < n +\frac{k^2 - k}2  \Longrightarrow \frac{k^2 + k}2 < n$

But $n$ is constant while $k$ is increasing, so eventually we will have a contradiction and $b_k \le k$. Therefore, the sequence of $a_i$s will become constant.

Solution 2

By the above, we have that

$b_{k+1} = \frac{b_k \cdot k + a_{k+1}}{k+1} = \left(\frac{k}{k+1}\right) \cdot b_k + \frac{a_{k+1}}{k+1}$

$\frac{k}{k+1} < 1$, and by definition, $\frac{a_{k+1}}{k+1} < 1$. Thus, $b_{k+1} < b_k + 1$. Also, both $b_k,\ b_{k+1}$ are integers, so $b_{k+1} \le b_k$. As the $b_k$s form a non-increasing sequence of positive integers, they must eventually become constant.

Therefore, $b_k = b_{k+1}$ for some sufficiently large value of $k$. Then $a_{k+1} = s_{k+1} - s_k = b_k(k + 1) - b_k(k) = b_k$, so eventually the sequence $a_k$ becomes constant.

Solution 3

Let $a_1=n$. Since $a_k\le k-1$, we have that $a_1+a_2+a_3+\hdots+a_n\le n+1+2+\hdots+n-1$.

Thus, $a_1+a_2+\hdots+a_n\le \frac{n(n+1)}{2}$.

Since $a_1+a_2+\hdots+a_n=nk$, for some integer $k$, we can keep adding $k$ to satisfy the conditions, provided that $k\le n$ because $a_n+1\le n$.

Because $k\le \frac{n+1}{2}\le n$, the sequence must eventually become constant.

See also

2007 USAMO (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions