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

Revision as of 10:43, 16 June 2009

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

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 2

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.

@@ Line 6: / Line 6: @@
 == Solution ==
-=== Solution 1 ===
-Define <math>s_k =\sum_{i=1}^{k} a_i</math>, and <math>b_k = \frac{s_k}{k}</math>. If <math>b_k \le k</math>, then for <math>k + 1</math>, <math>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}</math>. Note that <math>b_k</math> is a permissible value of <math> a_{k+1}</math> since <math>b_k \le k = (k+1)-1</math>: if we [[substitute]] <math>b_k</math> for <math> a_{k+1}</math>, we get <math>b_{k+1} = \frac{b_k (k+1)}{k+1} = b_k</math>, the unique value for <math> a_{k+1}</math>. So <math>b_k = b_{k+1} = b_{k+2} = \cdots</math>, from which if follows that the <math>a_k</math>s become constant.
-Now we must show that eventually <math>b_k \le k</math>. Suppose that <math>b_k > k</math> for all <math>k</math>. By definition, <math>\frac {s_k}{k} = b_k > k</math>, so <math>s_k > k^2</math>. Also, for <math>i>1</math>, each <math>a_i \le i-1</math> so
+=== Solution 1 ===
-<div style="text-align:center;"><math>k^2 <  s_k \le n + 1 + 2 + \cdots + (k-1) = n + \frac{k^2 - k}2</math> <br>
-<math>k^2 < n +\frac{k^2 - k}2  \Longrightarrow \frac{k^2 + k}2 < n</math></div>
-But <math>n</math> is constant while <math>k</math> is increasing, so eventually we will have a [[proof by contradiction|contradiction]] and <math>b_k \le k</math>. Therefore, the sequence of <math>a_i</math>s will become constant.
-=== Solution 2 ===
 By the above, we have that
@@ Line 25: / Line 16: @@
 Therefore, <math>b_k = b_{k+1}</math> for some sufficiently large value of <math>k</math>. Then <math>a_{k+1} = s_{k+1} - s_k = b_k(k + 1) - b_k(k) = b_k</math>, so eventually the sequence <math>a_k</math> becomes constant.
-=== Solution 3 ===
+=== Solution 2 ===
 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>.

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