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

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.

@@ Line 12: / Line 12: @@
 === 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.
+=== 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.
@@ Line 26: / Line 31: @@
 k &> b_k,
 \end{align*}</cmath>
-as desired. <math>\blacksquare</math>
+as desired.
+'''End Lemma'''
 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>.
+=== Solution 4 ===
+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.
+Let <math>m</math> be a positive integer such that <math>s_m = m(m - 1)</math>. We claim that
+<cmath>m - 1 = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \cdots.</cmath>
+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
+<cmath>0\leq a_{m+i} = m - 1\leq m + i - 1,</cmath>
+and yields
+<cmath>s_{m+i} = s_m + i(m - 1) = m(m - 1) + i(m - 1) = (m + i)(m - 1).</cmath>
+==Solution 5(like solution 2)==
+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:
+<cmath>\sum_{m=2}^{k} a_m =(q+1)k - p , k < n</cmath>
+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:
+<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>.
 {{alternate solutions}}

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