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

Revision as of 21:06, 30 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 $\displaystyle a_1 + a_2 + \cdots + a_k$ is divisible by $k$ . For instance, when $n=9$ the obtained sequence is $\displaystyle 9, 1, 2, 0, 3, 3, 3, \ldots$ . Prove that for any $n$ the sequence $\displaystyle a_1, a_2, a_3, \ldots$ eventually becomes constant.

Solution

Solution 1

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

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

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

But $\displaystyle n$ 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.

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

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

@@ Line 21: / Line 21: @@
 <div style="text-align:center;"><math>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}</math></div>
-<math>\frac{k}{k+1} < 1</math>, and by definition, <math>\frac{a_{k+1}}{k+1} < 1</math>. Thus, <math>\displaystyle b_{k+1} < b_k + 1</math>. Also, both <math>b_k,\ b_{k+1}</math> are integers, so <math>b_{k+1} \le b_k</math>. As the <math>\displaystyle b_k</math>s form a [[non-increasing]] sequence of positive integers, they must eventually become constant. Continue as above.
+<math>\frac{k}{k+1} < 1</math>, and by definition, <math>\frac{a_{k+1}}{k+1} < 1</math>. Thus, <math>\displaystyle b_{k+1} < b_k + 1</math>. Also, both <math>b_k,\ b_{k+1}</math> are integers, so <math>b_{k+1} \le b_k</math>. As the <math>\displaystyle b_k</math>s form a [[non-increasing]] sequence of positive integers, they must eventually become constant.
+Therefore, <math>\displaystyle b_k = b_{k+1}</math> for some sufficiently large value of <math>\displaystyle k</math>. Then <math>\displaystyle a_{k+1} = s_{k+1} - s_k = b_k(k + 1) - b_k(k) = b_k</math>, so eventually the sequence <math>\displaystyle a_k</math> becomes constant.
 == See also ==

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

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