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

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=== Solution 1 ===
 
=== 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>,
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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>,
 
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<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>
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<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.
 
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 ===
 
=== 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>. 
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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:
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'''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.
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''Proof:'' Choose any <math>k</math> such that <math>k^2 + 3k - 2 > 2n</math>. Then
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<cmath>\begin{align*}
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\frac{k^2 + 3k - 2}{2} &> n \\
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k^2 &> \frac{k^2 - 3k + 2}{2} + n \\
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k^2 &> (k-2) + (k-1) + \cdots + 1 + n \\
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k^2 &> a_{k-1} + a_{k-2} + \cdots + a_2 + a_1 \\
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k^2 &> S_k \\
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k &> \frac{S_k}{k} \\
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k &> b_k,
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\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.''
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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}}
 
{{alternate solutions}}

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.

See also

  • <url>viewtopic.php?t=145842 Discussion on AoPS/MathLinks</url>
2007 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions

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