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Difference between revisions of "2006 Canadian MO Problems/Problem 1"

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==Problem==
 
==Problem==
 
Let <math>f(n,k)</math> be the number of ways distributing <math>k</math> candies to <math>n</math> children so that each child receives at most two candies. For example, <math>f(3,7)=0</math>, <math>f(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.
 
Let <math>f(n,k)</math> be the number of ways distributing <math>k</math> candies to <math>n</math> children so that each child receives at most two candies. For example, <math>f(3,7)=0</math>, <math>f(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.
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apparently the proposers didnt get a closed form answer of this problem and somehow the previous solution here was wrong, this is standard roots of unity filter
 
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Let <cmath>f(n, k)</cmath> be the number of solutions <cmath>(x_1, x_2, \ldots, x_n)</cmath> to the equation
 
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<cmath>
 
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x_1 + x_2 + \cdots + x_n = k
 
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</cmath>
 
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where <cmath>x_i \in \{0, 1, 2\}</cmath> for <cmath>1 \le i \le n</cmath>. Equivalently, <cmath>f(n, k)</cmath> is the coefficient of <cmath>x^k</cmath> in the generating function
 
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<cmath>
 
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A(x) = (1 + x + x^2)^n.
 
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</cmath>
 
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Define
 
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<cmath>
 
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A(x) = \sum_{k=0}^{\infty} f(n, k) x^k.
 
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</cmath>
 
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Let <cmath>\omega \ne 1</cmath> be a primitive third root of unity. Then
 
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<cmath>
 
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\frac{1}{3} \left[ A(x) + \omega^2 A(\omega x) + \omega A(\omega^2 x) \right] = f(n,1)x + f(n,4)x^4 + f(n,7)x^7 + \cdots.
 
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</cmath>
 
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Plugging in <cmath>x = 1</cmath>, we obtain
 
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<cmath>
 
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\frac{1}{3} \left[ A(1) + \omega^2 A(\omega) + \omega A(\omega^2) \right] = f(n,1) + f(n,4) + f(n,7) + \cdots.
 
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</cmath>
 
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We compute:
 
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<cmath>
 
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A(1) = 3^n, \quad A(\omega) = 0, \quad A(\omega^2) = 0,
 
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</cmath>
 
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so the desired sum is
 
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<cmath>
 
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\frac{1}{3} \cdot 3^n = 3^{n-1}.
 
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</cmath>
 
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Thus, the final answer is <cmath>3^{2005}</cmath>.
 
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~Ishan
 
  
 
==See also==
 
==See also==

Revision as of 12:15, 20 May 2025

Problem

Let $f(n,k)$ be the number of ways distributing $k$ candies to $n$ children so that each child receives at most two candies. For example, $f(3,7)=0$, $f(3,6)=1$, and $f(3,4)=6$. Evaluate $f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)$.

See also

2006 Canadian MO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 Followed by
Problem 2
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