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

Revision as of 21:30, 6 August 2014

Problem

(Sam Vandervelde) 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.

Solutions

Solution 1

Let $S_k = a_1 + a_2 + \cdots + a_k$ and $b_k = \frac{S_k}{k}$ . Thus, because $S_{k+1} = S_k + a_{k+1}$ , $b_{k+1} = \frac{b_k \cdot k + a_{k+1}}{k+1} = \left(\frac{k}{k+1}\right) \cdot b_k + \frac{a_{k+1}}{k+1}$ $\frac{k}{k+1} < 1$ , and by definition, $\frac{a_{k+1}}{k+1} < 1$ . Thus, $b_{k+1} < b_k + 1$ . Also, both $b_k$ and $b_{k+1}$ 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

Define $S_k = a_1 + a_2 + \cdots + a_k$ , and $b_k = \frac{S_k}{k}$ . By the problem hypothesis, $b_k$ is an integer valued sequence.

Lemma: There exists a $k$ such that $b_k < k$ .

Proof: Choose any $k$ such that $k^2 + 3k - 2 > 2n$ . Then $\begin{align*} \frac{k^2 + 3k - 2}{2} &> n \\ k^2 &> \frac{k^2 - 3k + 2}{2} + n \\ k^2 &> (k-2) + (k-1) + \cdots + 1 + n \\ k^2 &> a_{k-1} + a_{k-2} + \cdots + a_2 + a_1 \\ k^2 &> S_k \\ k &> \frac{S_k}{k} \\ k &> b_k, \end{align*}$ as desired. $\blacksquare$

Let $k$ be the smallest $k$ such that $b_k < k$ . Then $b_k = m < k$ , and $S_k = km$ . To make $b_{k+1}$ an integer, $S_{k+1} = S_k + a_{k+1}$ must be divisible by $k+1$ . Thus, because $km + m$ is divisible by $k+1$ , $a_{k+1} \equiv m \pmod{k+1}$ , and, because $0 \le a_{k+1} < k$ , $a_{k+1} = m$ . Then $b_{k+1} = \frac{(k+1)m}{k+1} = m$ as well. Repeating the same process using $k+1$ instead of $k$ gives $a_{k+2} = m$ , and an easy induction can prove that for all $N > k+1$ , $a_N = m$ . Thus, $a_k$ becomes a constant function for arbitrarily large values of $k$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 5: / Line 5: @@
 === Solution 1 ===
-Let <math>S_k = a_1 + a_2 + ... + a_k</math> and <math>b_k = \frac{S_k}{k}</math>. Thus, because <math>S_{k+1} = S_k + a_{k+1}</math>,
+Let <math>S_k = a_1 + a_2 + \cdots + a_k</math> and <math>b_k = \frac{S_k}{k}</math>. Thus, because <math>S_{k+1} = S_k + a_{k+1}</math>,
+<cmath>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}</cmath>
-<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>b_{k+1} < b_k + 1</math>. Also, both <math>b_k</math> and <math>b_{k+1}</math> are integers, so <math>b_{k+1} \le b_k</math>. As the <math>b_k</math>'s form a [[non-increasing]] sequence of positive integers, they must eventually become constant.
-<math>\frac{k}{k+1} < 1</math>, and by definition, <math>\frac{a_{k+1}}{k+1} < 1</math>. Thus, <math>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>b_k</math>s form a [[non-increasing]] sequence of positive integers, they must eventually become constant.
 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 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>.
+Define <math>S_k = a_1 + a_2 + \cdots + a_k</math>, and <math>b_k = \frac{S_k}{k}</math>. By the problem hypothesis, <math>b_k</math> is an integer valued sequence.
-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>.
-Because <math>k\le \frac{n+1}{2}\le n</math>, the sequence must eventually become constant.
-===Solution 3===
-Define <math>S_k = a_1 + a_2 + ... + a_k</math>, and <math>b_k = \frac{S_k}{k}</math>. By the problem hypothesis, <math>b_k</math> is an integer valued sequence.
-''Lemma:'' The exists a <math>k</math> such that <math>b_k < k</math>.
-Proof: Choose any <math>k</math> such that <math>k^2 + 3k - 2 > 2n</math>. Then:
+'''Lemma:''' There exists a <math>k</math> such that <math>b_k < k</math>.
-<cmath>\frac{k^2 + 3k - 2}{2} > n</cmath>
-<cmath>k^2 > \frac{k^2 - 3k + 2}{2} + n</cmath>
-<cmath>k^2 > (k-2) + (k-1) + ... + 1 + n</cmath>
-<cmath>k^2 > a_{k-1} + a_{k-2} + ... + a_2 + a_1</cmath>
-<cmath>k^2 > S_k</cmath>
-<cmath>k > \frac{S_k}{k}</cmath>
-<cmath>k > b_k,</cmath>
-as desired.
-Let k be the smallest k such that <math>b_k < k</math>. Then <math>b_k = m < k</math>, and <math>S_k = km</math>. To make <math>b_{k+1}</math> an integer, <math>S_{k+1} = S_k + a_{k+1}</math> must be divisible by <math>k+1</math>. Thus, because <math>km + m</math> is divisible by <math>k+1</math>, <math>a_{k+1} \equiv m \mod (k+1)</math>, and, because <math>0 \le a_{k+1} < k</math>, <math>a_{k+1} = m</math>. Then <math>b_{k+1} = \frac{(k+1)m}{k+1} = m</math> as well. Repeating the same process using <math>k+1</math> instead of <math>k</math> gives <math>a_{k+2} = m</math>, and an easy induction can prove that for all <math>N > k+1</math>, <math>a_N = m</math>. Thus, <math>a_k</math> becomes a constant function for arbitrarily large values of k.
+''Proof:'' Choose any <math>k</math> such that <math>k^2 + 3k - 2 > 2n</math>. Then
+<cmath>\begin{align*}
+\frac{k^2 + 3k - 2}{2} &> n \
+k^2 &> \frac{k^2 - 3k + 2}{2} + n \
+k^2 &> (k-2) + (k-1) + \cdots + 1 + n \
+k^2 &> a_{k-1} + a_{k-2} + \cdots + a_2 + a_1 \
+k^2 &> S_k \
+k &> \frac{S_k}{k} \
+k &> b_k,
+\end{align*}</cmath>
+as desired. <math>\blacksquare</math>
-''Note: This solution is a formalization of the second solution. Also, the lemma could have been simplified if I chose k = n, which is exactly the second solution's thought process.''
+Let <math>k</math> be the smallest <math>k</math> such that <math>b_k < k</math>. Then <math>b_k = m < k</math>, and <math>S_k = km</math>. To make <math>b_{k+1}</math> an integer, <math>S_{k+1} = S_k + a_{k+1}</math> must be divisible by <math>k+1</math>. Thus, because <math>km + m</math> is divisible by <math>k+1</math>, <math>a_{k+1} \equiv m \pmod{k+1}</math>, and, because <math>0 \le a_{k+1} < k</math>, <math>a_{k+1} = m</math>. Then <math>b_{k+1} = \frac{(k+1)m}{k+1} = m</math> as well. Repeating the same process using <math>k+1</math> instead of <math>k</math> gives <math>a_{k+2} = m</math>, and an easy induction can prove that for all <math>N > k+1</math>, <math>a_N = m</math>. Thus, <math>a_k</math> becomes a constant function for arbitrarily large values of <math>k</math>.
 {{alternate solutions}}

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

Page

Toolbox

Search