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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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==Solution==
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<cmath>
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<math>x_1 + \cdots + x_{2006} = k</math>
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x_1 + x_2 + \cdots + x_n = k
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</cmath>
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<math>x_i \in \{0,1,2\}</math>
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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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<math>\therefore</math> generating function for one student is <math>f(x) = (x^0 + x^1 + x^2)</math>
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A(x) = (1 + x + x^2)^n.
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</cmath>
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generating function for <math>n</math> many students is <math>f(x)^n = (x^0 + x^1 + x^2)^n</math>
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Define
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<cmath>
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<math>\therefore</math> generating function for 2006 many students is <math>(x^0 + x^1 + x^2)^{2006}</math>
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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 <math>g(x) = (x^0 + x^1 + x^2)^{2006} = (1 + x + x^2)^{2006}</math>
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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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<math>g(x) = \sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k</math>
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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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<math>\sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k = g(x)</math>
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Plugging in <cmath>x = 1</cmath>, we obtain
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<cmath>
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<math>\therefore</math> coefficient of <math>x^{3k+1}</math> is <math>f(2006, 3k+1)</math>, \quad <math>k \in \{0,1,\dots\}</math>
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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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Take <math>h(x) = g(x) x^{-1} = \frac{(1 + x + x^2)^{2006}}{x}</math>
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We compute:
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<cmath>
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Then <math>\sum_{k=1}^{1337} f(2006, 3k+1)</math> is the sum of the coefficients of <math>x^{3m},\ m \in \mathbb{N}</math> in <math>h(x)</math>
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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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<math>\omega = e^{2\pi i/3}</math> \quad (third root of unity)
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so the desired sum is
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<cmath>
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and since <math>1 + \omega + \omega^2 = 0</math>
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\frac{1}{3} \cdot 3^n = 3^{n-1}.
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</cmath>
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the other coefficients vanish when we add up
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Thus, the final answer is <cmath>3^{2005}</cmath>.
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<math>g(\omega^2) + g(\omega) + g(1) = \frac{(1 + \omega + \omega^2)^{2006}}{\omega} + \frac{(1 + \omega^2 + \omega^4)^{2006}}{\omega^2} + \frac{3^{2006}}{1}</math>
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<math>= \frac{3^{2006}}{1} + \frac{0}{\omega^2} + \frac{0}{\omega} = 3^{2006}</math>
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and since we added up the same coefficient thrice, we need to divide it by <math>3</math>
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<math>\therefore \sum_{k=1}^{1337} f(2006, 3k+1) = \frac{3^{2006}}{3} = 3^{2005}</math>
 
~Ishan
 
~Ishan
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==See also==
 
==See also==
 
*[[2006 Canadian MO]]
 
*[[2006 Canadian MO]]
  
 
{{CanadaMO box|year=2006|before=First question|num-a=2}}
 
{{CanadaMO box|year=2006|before=First question|num-a=2}}

Latest revision as of 12:22, 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 $f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)$.

Solution

$x_1 + \cdots + x_{2006} = k$

$x_i \in \{0,1,2\}$

$\therefore$ generating function for one student is $f(x) = (x^0 + x^1 + x^2)$

generating function for $n$ many students is $f(x)^n = (x^0 + x^1 + x^2)^n$

$\therefore$ generating function for 2006 many students is $(x^0 + x^1 + x^2)^{2006}$

Let $g(x) = (x^0 + x^1 + x^2)^{2006} = (1 + x + x^2)^{2006}$

$g(x) = \sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k$

$\sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k = g(x)$

$\therefore$ coefficient of $x^{3k+1}$ is $f(2006, 3k+1)$, \quad $k \in \{0,1,\dots\}$

Take $h(x) = g(x) x^{-1} = \frac{(1 + x + x^2)^{2006}}{x}$

Then $\sum_{k=1}^{1337} f(2006, 3k+1)$ is the sum of the coefficients of $x^{3m},\ m \in \mathbb{N}$ in $h(x)$

$\omega = e^{2\pi i/3}$ \quad (third root of unity)

and since $1 + \omega + \omega^2 = 0$

the other coefficients vanish when we add up

$g(\omega^2) + g(\omega) + g(1) = \frac{(1 + \omega + \omega^2)^{2006}}{\omega} + \frac{(1 + \omega^2 + \omega^4)^{2006}}{\omega^2} + \frac{3^{2006}}{1}$

$= \frac{3^{2006}}{1} + \frac{0}{\omega^2} + \frac{0}{\omega} = 3^{2006}$

and since we added up the same coefficient thrice, we need to divide it by $3$

$\therefore \sum_{k=1}^{1337} f(2006, 3k+1) = \frac{3^{2006}}{3} = 3^{2005}$ ~Ishan

See also

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