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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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==Solution==
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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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<math>f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies.} </math>
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<cmath>
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x_1 + x_2 + \cdots + x_n = k
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<math> \text{This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i.</math>
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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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<math>\text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n. </math>
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<cmath>
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A(x) = (1 + x + x^2)^n.
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<math>\text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}. </math>
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</cmath>
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Define
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<math> \text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m. </math>
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<cmath>
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A(x) = \sum_{k=0}^{\infty} f(n, k) x^k.
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<math>\text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}. </math>
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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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<math> \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. </math>
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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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<math>\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. </math>
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</cmath>
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Plugging in <cmath>x = 1</cmath>, we obtain
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<math>\text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. </math>
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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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<math>\text{ Therefore, the final answer is } \boxed{2^{2005}}. </math>
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</cmath>
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We compute:
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~sitar
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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 12:08, 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 $f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)$. 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 Let \[f(n, k)\] be the number of solutions \[(x_1, x_2, \ldots, x_n)\] to the equation \[x_1 + x_2 + \cdots + x_n = k\] where \[x_i \in \{0, 1, 2\}\] for \[1 \le i \le n\]. Equivalently, \[f(n, k)\] is the coefficient of \[x^k\] in the generating function \[A(x) = (1 + x + x^2)^n.\] Define \[A(x) = \sum_{k=0}^{\infty} f(n, k) x^k.\] Let \[\omega \ne 1\] be a primitive third root of unity. Then \[\frac{1}{3} \left[ A(x) + \omega^2 A(\omega x) + \omega A(\omega^2 x) \right] = f(n,1)x + f(n,4)x^4 + f(n,7)x^7 + \cdots.\] Plugging in \[x = 1\], we obtain \[\frac{1}{3} \left[ A(1) + \omega^2 A(\omega) + \omega A(\omega^2) \right] = f(n,1) + f(n,4) + f(n,7) + \cdots.\] We compute: \[A(1) = 3^n, \quad A(\omega) = 0, \quad A(\omega^2) = 0,\] so the desired sum is \[\frac{1}{3} \cdot 3^n = 3^{n-1}.\] Thus, the final answer is \[3^{2005}\]. ~Ishan

See also

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