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

Revision as of 01:04, 9 October 2025

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

Let $S_k =\sum_{i=1}^k a_i$ , where $S_k \equiv 0 \pmod{k}$ and $0 \leq a_k < k$ . Define $m_k = S_k / k$ , with $m_1 = n$ .

First, for $k > 1$ , since $S_{k-1} = (k-1) \cdot m_{k-1}$ and $S_k = S_{k-1} + a_k$ , we need $S_k \equiv 0 \pmod{k}$ . Set $a_k \equiv m_{k-1} \pmod{k}$ . Then,

$S_k \bmod k \equiv \bigl((k-1) \cdot m_{k-1} + (m_{k-1} \bmod k)\bigr) \pmod{k}$ . Since $(k-1) \cdot m_{k-1} \equiv -m_{k-1} \pmod{k}$ , and $-m_{k-1} + (m_{k-1} \bmod k) \equiv 0 \pmod{k}$ , this holds.

Next, compute $m_k = S_k / k = \bigl((k-1) \cdot m_{k-1} + a_k\bigr) / k$ . If $m_{k-1} \geq k$ , let $m_{k-1} = qk + r$ with $0 \leq r < k$ , so $a_k = r$ , and $m_k = \frac{(k-1) \cdot m_{k-1} + r}{k} < m_{k-1}$ , so $m_k < m_{k-1}$ . Moreover, so long as $m_{k-1} \geq k$ then $m_{i}$ is strictly decreasing while $k$ is strictly increasing, until $m_{i} \leq k$ , in which case the sequence terminates as $m_{k-1} = m_{k}$

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 50: / Line 50: @@
 and yields
 <cmath>s_{m+i} = s_m + i(m - 1) = m(m - 1) + i(m - 1) = (m + i)(m - 1).</cmath>
+===Solution 5===
+Let <math>S_k =\sum_{i=1}^k a_i</math>, where <math>S_k \equiv 0 \pmod{k}</math> and <math>0 \leq a_k < k</math>. Define <math>m_k = S_k / k</math>, with <math>m_1 = n</math>.
+First, for <math>k > 1</math>, since <math>S_{k-1} = (k-1) \cdot m_{k-1}</math> and <math>S_k = S_{k-1} + a_k</math>, we need <math>S_k \equiv 0 \pmod{k}</math>. Set <math>a_k \equiv m_{k-1} \pmod{k}</math>. Then,
+<math>S_k \bmod k \equiv \bigl((k-1) \cdot m_{k-1} + (m_{k-1} \bmod k)\bigr) \pmod{k}</math>.
+Since <math>(k-1) \cdot m_{k-1} \equiv -m_{k-1} \pmod{k}</math>, and <math>-m_{k-1} + (m_{k-1} \bmod k) \equiv 0 \pmod{k}</math>, this holds.
+Next, compute <math>m_k = S_k / k = \bigl((k-1) \cdot m_{k-1} + a_k\bigr) / k</math>.
+If <math>m_{k-1} \geq k</math>, let <math>m_{k-1} = qk + r</math> with <math>0 \leq r < k</math>, so <math>a_k = r</math>, and
+<math>m_k = \frac{(k-1) \cdot m_{k-1} + r}{k} < m_{k-1}</math>,
+so <math>m_k < m_{k-1}</math>. Moreover, so long as <math>m_{k-1} \geq k</math> then <math>m_{i}</math> is strictly decreasing while <math>k</math> is strictly increasing, until <math>m_{i} \leq k</math>, in which case the sequence terminates as <math>m_{k-1} = m_{k}</math>
 ==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:

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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