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

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== Problem ==
 
== Problem ==
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(''Sam Vandervelde'') Let <math>n</math> be a [[positive]] [[integer]].  Define a [[sequence]] by setting <math>a_1 = n</math> and, for each <math>k>1</math>, letting <math>a_k</math> be the unique integer in the range <math>0 \le a_k \le k-1</math> for which <math>a_1 + a_2 + \cdots + a_k</math> is [[divisible]] by <math>k</math>.  For instance, when <math>n=9</math> the obtained sequence is <math>9, 1, 2, 0, 3, 3, 3, \ldots</math>.  Prove that for any <math>n</math> the sequence <math>a_1, a_2, a_3, \ldots</math> eventually becomes [[constant]].
  
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Let <math>n</math> be a [[positive]] [[integer]].  Define a [[sequence]] by setting <math>a_1 = n</math> and, for each <math>k>1</math>, letting <math>a_k</math> be the unique integer in the range <math>0 \le a_k \le k-1</math> for which <math>a_1 + a_2 + \cdots + a_k</math> is [[divisible]] by <math>k</math>.  For instance, when <math>n=9</math> the obtained sequence is <math>9, 1, 2, 0, 3, 3, 3, \ldots</math>.  Prove that for any <math>n</math> the sequence <math>a_1, a_2, a_3, \ldots</math> eventually becomes [[constant]].
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== Solutions ==
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__TOC__
 
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== Solution ==
 
  
 
=== Solution 1 ===
 
=== Solution 1 ===
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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>,
 +
<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>
 +
<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.
  
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<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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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.
  
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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,\ 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.  
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=== Solution 2 ===
 +
Let <math>a_1 = n</math>. Since <math>a_k\leq k - 1</math>, we have that
 +
<cmath>a_1 + a_2 + \cdots + a_n\leq n + 1 + 2 + \cdots + n - 1 = \frac{n(n + 1)}{2}.</cmath>
 +
Since <math>a_1 + a_2 + \cdots + 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\leq n</math>. This is true since <math>k\leq\frac{n + 1}{2}\leq n</math>, so the sequence 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.
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=== Solution 3 ===
 +
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.
  
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=== Solution 2 ===
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'''Lemma:''' There exists a <math>k</math> such that <math>b_k < k</math>.
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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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Thus, <math>a_1+a_2+\hdots+a_n\le \frac{n(n+1)}{2}</math>.
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''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} \\
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k &> b_k,
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\end{align*}</cmath>
 +
as desired.
  
−
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>.
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'''End Lemma'''
  
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Because <math>k\le \frac{n+1}{2}\le n</math>, the sequence must eventually become constant.
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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>.
  
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===Solution 3===
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=== Solution 4 ===
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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.
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For <math>k\geq 1</math>, let
 +
<cmath>s_k = a_1 + a_2 + \cdots + a_k.</cmath>
 +
We claim that for some <math>m</math> we have <math>s_m = m(m - 1)</math>. To this end, consider the sequence which computes the differences between <math>s_k</math> and <math>k(k - 1)</math>, i.e., whose <math>k</math>-th term is <math>s_k - k(k - 1)</math>. Note that the first term of this sequence is positive (it is equal to <math>n</math>) and that its terms are strictly decreasing since
 +
<cmath>(s_k - k(k - 1)) - (s_{k+1} - (k + 1)k) = 2k - a_{k+1}\geq 2k - k = k\geq 1.</cmath>
 +
Further, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that <math>s_k > k(k - 1)</math> and <math>s_{k+1} < (k + 1)k</math>. Since <math>s_k</math> and <math>s_{k+1}</math> are divisible by <math>k</math> and <math>k + 1</math>, respectively, we can tighten the above inequalities to <math>s_k\geq k^2</math> and <math>s_{k+1}\leq (k + 1)(k - 1) = k^2 - 1</math>. But this would imply that <math>s_k > s_{k+1}</math>, a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.
  
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''Lemma:'' The exists a <math>k</math> such that <math>b_k < k</math>.
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Let <math>m</math> be a positive integer such that <math>s_m = m(m - 1)</math>. We claim that
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<cmath>m - 1 = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \cdots.</cmath>
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Proof: Choose any <math>k</math> such that <math>k^2 + 3k - 2 > 2n</math>. Then:
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This follows from the fact that the sequence <math>a_1, a_2, a_3, \ldots</math> is uniquely determined and choosing <math>a_{m+i} = m - 1</math>, for <math>i\geq 1</math>, satisfies the range condition
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<cmath>\frac{k^2 + 3k - 2}{2} > n</cmath>
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<cmath>0\leq a_{m+i} = m - 1\leq m + i - 1,</cmath>
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<cmath>k^2 > \frac{k^2 - 3k + 2}{2} + n</cmath>
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and yields
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<cmath>k^2 > (k-2) + (k-1) + ... + 1 + n</cmath>
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<cmath>s_{m+i} = s_m + i(m - 1) = m(m - 1) + i(m - 1) = (m + i)(m - 1).</cmath>
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<cmath>k^2 > a_{k-1} + a_{k-2} + ... + a_2 + a_1</cmath>
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==Solution 5(like solution 2)==
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<cmath>k^2 > S_k</cmath>
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First, we may make an observation and say that for <math>n \equiv p (mod k)</math>, <math>\sum_{m=2}^{k} a_m \equiv k-p (mod k)</math> must occur for the whole sum to be divisible by <math>k</math>. Thus, the following is apparent:
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<cmath>k > \frac{S_k}{k}</cmath>
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<cmath>\sum_{m=2}^{k} a_m =(q+1)k - p , k < n</cmath>
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<cmath>k > b_k,</cmath>
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Then, we may make another observation that when <math>n=k</math>, the sum also has to be divisible by n. We may then explore when n=k:
−
as desired.
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<cmath> \sum_{m=2}^{n} a_m \equiv 0 (mod n), \sum_{m=2}^{n} a_m = rn</cmath>
 +
and
 +
<cmath>a_n = \sum_{m=2}^{n} a_m - \sum_{m=2}^{n-1} a_m</cmath>
 +
Then,
 +
<cmath>a_n = rn - \sum_{m=2}^{n-1} a_m \le n-1</cmath>
 +
Also,
 +
<cmath>a_{n+s} = r, r \le n</cmath> for <math>s \ge 1</math>. This is because:
 +
<cmath>\sum_{m=2}^{n} a_m + r = rn + r = r(n+1)</cmath>
 +
This must be true since <math>r(n+1)</math> will be divisible by <math>n+1</math> and <math>k=n+1</math>, we may then generalize this to all <math>r(n+s), s \in \mathbb{Z}, s \ge 1</math>
 +
<cmath>rn + sr= r(n+s), \frac{r(n+s)}{r} = n + s = \frac{\sum_{m=2}^{n+s} a_m}{r}</cmath>
 +
Thus, we may say that the sequence <math>a_1,a_2,...a_k</math> must converge to some integer value <math>r \le n</math> when <math>k \ge n + 1</math>.
  
−
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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''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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{{alternate solutions}}
  
 
== See also ==
 
== See also ==
 +
* <url>viewtopic.php?t=145842 Discussion on AoPS/MathLinks</url>
  
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{{USAMO newbox|year=2007|before=First question|num-a=2}}
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{{USAMO newbox|year=2007|before=First Question|num-a=2}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:06, 23 August 2023

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

Let $a_1 = n$. Since $a_k\leq k - 1$, we have that \[a_1 + a_2 + \cdots + a_n\leq n + 1 + 2 + \cdots + n - 1 = \frac{n(n + 1)}{2}.\] Since $a_1 + a_2 + \cdots + a_n = nk$ for some integer $k$, we can keep adding $k$ to satisfy the conditions, provided that $k\leq n$. This is true since $k\leq\frac{n + 1}{2}\leq n$, so the sequence must eventually become constant.

Solution 3

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.

End Lemma

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 $k$.

Solution 4

For $k\geq 1$, let \[s_k = a_1 + a_2 + \cdots + a_k.\] We claim that for some $m$ we have $s_m = m(m - 1)$. To this end, consider the sequence which computes the differences between $s_k$ and $k(k - 1)$, i.e., whose $k$-th term is $s_k - k(k - 1)$. Note that the first term of this sequence is positive (it is equal to $n$) and that its terms are strictly decreasing since \[(s_k - k(k - 1)) - (s_{k+1} - (k + 1)k) = 2k - a_{k+1}\geq 2k - k = k\geq 1.\] Further, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that $s_k > k(k - 1)$ and $s_{k+1} < (k + 1)k$. Since $s_k$ and $s_{k+1}$ are divisible by $k$ and $k + 1$, respectively, we can tighten the above inequalities to $s_k\geq k^2$ and $s_{k+1}\leq (k + 1)(k - 1) = k^2 - 1$. But this would imply that $s_k > s_{k+1}$, a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.

Let $m$ be a positive integer such that $s_m = m(m - 1)$. We claim that \[m - 1 = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \cdots.\] This follows from the fact that the sequence $a_1, a_2, a_3, \ldots$ is uniquely determined and choosing $a_{m+i} = m - 1$, for $i\geq 1$, satisfies the range condition \[0\leq a_{m+i} = m - 1\leq m + i - 1,\] and yields \[s_{m+i} = s_m + i(m - 1) = m(m - 1) + i(m - 1) = (m + i)(m - 1).\]

Solution 5(like solution 2)

First, we may make an observation and say that for $n \equiv p (mod k)$, $\sum_{m=2}^{k} a_m \equiv k-p (mod k)$ must occur for the whole sum to be divisible by $k$. Thus, the following is apparent: \[\sum_{m=2}^{k} a_m =(q+1)k - p , k < n\] Then, we may make another observation that when $n=k$, the sum also has to be divisible by n. We may then explore when n=k: \[\sum_{m=2}^{n} a_m \equiv 0 (mod n), \sum_{m=2}^{n} a_m = rn\] and \[a_n = \sum_{m=2}^{n} a_m - \sum_{m=2}^{n-1} a_m\] Then, \[a_n = rn - \sum_{m=2}^{n-1} a_m \le n-1\] Also, \[a_{n+s} = r, r \le n\] for $s \ge 1$. This is because: \[\sum_{m=2}^{n} a_m + r = rn + r = r(n+1)\] This must be true since $r(n+1)$ will be divisible by $n+1$ and $k=n+1$, we may then generalize this to all $r(n+s), s \in \mathbb{Z}, s \ge 1$ \[rn + sr= r(n+s), \frac{r(n+s)}{r} = n + s = \frac{\sum_{m=2}^{n+s} a_m}{r}\] Thus, we may say that the sequence $a_1,a_2,...a_k$ must converge to some integer value $r \le n$ when $k \ge n + 1$.


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 (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
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