Difference between revisions of "2007 USAMO Problems/Problem 1"
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k &> b_k, | k &> b_k, | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | as desired. | + | 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>. | 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}} | {{alternate solutions}} |
Latest revision as of 19:06, 23 August 2023
Contents
Problem
(Sam Vandervelde) Let be a positive integer. Define a sequence by setting and, for each , letting be the unique integer in the range for which is divisible by . For instance, when the obtained sequence is . Prove that for any the sequence eventually becomes constant.
Solutions
Solution 1
Let and . Thus, because , , and by definition, . Thus, . Also, both and are integers, so . As the 's form a non-increasing sequence of positive integers, they must eventually become constant.
Therefore, for some sufficiently large value of . Then , so eventually the sequence becomes constant.
Solution 2
Let . Since , we have that Since for some integer , we can keep adding to satisfy the conditions, provided that . This is true since , so the sequence must eventually become constant.
Solution 3
Define , and . By the problem hypothesis, is an integer valued sequence.
Lemma: There exists a such that .
Proof: Choose any such that . Then as desired.
End Lemma
Let be the smallest such that . Then , and . To make an integer, must be divisible by . Thus, because is divisible by , , and, because , . Then as well. Repeating the same process using instead of gives , and an easy induction can prove that for all , . Thus, becomes a constant function for arbitrarily large values of .
Solution 4
For , let We claim that for some we have . To this end, consider the sequence which computes the differences between and , i.e., whose -th term is . Note that the first term of this sequence is positive (it is equal to ) and that its terms are strictly decreasing since Further, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that and . Since and are divisible by and , respectively, we can tighten the above inequalities to and . But this would imply that , a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.
Let be a positive integer such that . We claim that This follows from the fact that the sequence is uniquely determined and choosing , for , satisfies the range condition and yields
Solution 5(like solution 2)
First, we may make an observation and say that for , must occur for the whole sum to be divisible by . Thus, the following is apparent: Then, we may make another observation that when , the sum also has to be divisible by n. We may then explore when n=k: and Then, Also, for . This is because: This must be true since will be divisible by and , we may then generalize this to all Thus, we may say that the sequence must converge to some integer value when .
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 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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