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Difference between revisions of "2001 IMO Shortlist Problems/C2"

(I now know what a congruence class is.)
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== Solution ==
 
== Solution ==
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We shall prove this by contradiction. Assume that for some <math>n</math>-tuple of <math>c_i</math> there does not exist two permutations <math>a</math> and <math>b</math> of <math>\{ 1, 2, \ldots,n\}</math> such that <math>n!|S(a)-S(b)</math>. Note that there are <math>n!</math> distinct permutations of <math>\{1, 2, \ldots,n\}</math>, which means there are <math>n!</math> distinct sums <math>S(a)</math>. Since no two of them are congruent modulo <math>n!</math>, we have that for some <math>n</math>-tuple of <math>c_i</math> there exists a unique permutation <math>x</math> such that <math>S(x)\equiv y\bmod{n}</math> for all integers <math>0\leq y\leq n-1</math>. This means that the sum of every <math>S(a)</math> is congruent to <math>\sum_{i=0}^{n!-1}=\frac{(n!-1)n!}{2}</math>. However, that same sum is congruent to
+
We shall prove this by contradiction. Assume that for some <math>n</math>-tuple of <math>c_i</math> there does not exist two permutations <math>a</math> and <math>b</math> of <math>s_n=\{ 1, 2, \ldots,n\}</math> such that <math>n!|S(a)-S(b)</math>. Note that there are <math>n!</math> distinct permutations of <math>s_n</math>, which means there are <math>n!</math> distinct sums <math>S(a)</math>. Because no two of them are congruent modulo <math>n!</math>, we have that for some <math>n</math>-tuple of <math>c_i</math> there exists a unique permutation <math>x</math> of <math>s_n</math> such that <math>S(x)\equiv y\bmod{n}</math> for all integers <math>0\leq y\leq n-1</math>. This means that the sum of every <math>S(a)</math> is congruent to <math>\sum_{i=0}^{n!-1}=\frac{(n!-1)n!}{2}</math>. However, that same sum is congruent to
  
 
<cmath>\sum_{i=1}^{n}\left(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}</cmath>
 
<cmath>\sum_{i=1}^{n}\left(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}</cmath>

Latest revision as of 08:55, 31 August 2011

Problem

Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i = 1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a) - S(b)$.

Solution

We shall prove this by contradiction. Assume that for some $n$-tuple of $c_i$ there does not exist two permutations $a$ and $b$ of $s_n=\{ 1, 2, \ldots,n\}$ such that $n!|S(a)-S(b)$. Note that there are $n!$ distinct permutations of $s_n$, which means there are $n!$ distinct sums $S(a)$. Because no two of them are congruent modulo $n!$, we have that for some $n$-tuple of $c_i$ there exists a unique permutation $x$ of $s_n$ such that $S(x)\equiv y\bmod{n}$ for all integers $0\leq y\leq n-1$. This means that the sum of every $S(a)$ is congruent to $\sum_{i=0}^{n!-1}=\frac{(n!-1)n!}{2}$. However, that same sum is congruent to

\[\sum_{i=1}^{n}\left(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}\]

Note that $2|n+1$, so $n!|\frac{(n+1)!}{2}$, so $n!$ divides the sum. However, the sum is also congruent to $\frac{(n!-1)n!}{2}$ modulo $n!$, and $n!-1$ is odd, so $n!$ couldn't possibly divide the sum. This leads to a contradiction, so our previous assumption must be false. This proves the problem statement.

Resources

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