Difference between revisions of "2001 IMO Problems/Problem 4"

(New page: Let n1, n2, ... , nm be integers where m is odd. Let x = (x1, ... , xm) denote a permutation of the integers 1, 2, ... , m. Let f(x) = x1n1 + x2n2 + ... + xmnm. Show that for some distinct...)
 
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Let n1, n2, ... , nm be integers where m is odd. Let x = (x1, ... , xm) denote a permutation of the integers 1, 2, ... , m. Let f(x) = x1n1 + x2n2 + ... + xmnm. Show that for some distinct permutations a, b the difference f(a) - f(b) is a multiple of m!.
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=Problem==
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Let <math>n_1, n_2, \dots , n_m</math> be integers where <math>m</math> is odd. Let <math>x = (x_1, \dots , x_m)</math> denote a permutation of the integers <math>1, 2, \cdots , m</math>. Let <math>f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m</math>. Show that for some distinct permutations <math>a</math>, <math>b</math> the difference <math>f(a) - f(b)</math> is a multiple of <math>m!</math>.
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==Solution==
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{{solution}}
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==See also==
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[[Category:Olympiad Combinatorics Problems]]

Revision as of 10:28, 15 October 2008

Problem=

Let $n_1, n_2, \dots , n_m$ be integers where $m$ is odd. Let $x = (x_1, \dots , x_m)$ denote a permutation of the integers $1, 2, \cdots , m$. Let $f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m$. Show that for some distinct permutations $a$, $b$ the difference $f(a) - f(b)$ is a multiple of $m!$.

Solution

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See also