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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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<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. 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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==Solution==
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<math>x_1 + \cdots + x_{2006} = k</math>
  
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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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<math>x_i \in \{0,1,2\}</math>
  
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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}. \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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<math>\therefore</math> generating function for one student is <math>f(x) = (x^0 + x^1 + x^2)</math>
  
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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}. \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. </math>
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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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<math>\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. \text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. \text{ Therefore, the final answer is } \boxed{2^{2005}}. </math>
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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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Let <math>g(x) = (x^0 + x^1 + x^2)^{2006} = (1 + x + x^2)^{2006}</math>
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<math>g(x) = \sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k</math>
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<math>\sum_{k=0}^{2 \cdot 2006} f(2006, k) x^k = g(x)</math>
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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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Take <math>h(x) = g(x) x^{-1} = \frac{(1 + x + x^2)^{2006}}{x}</math>
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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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<math>\omega = e^{2\pi i/3}</math> \quad (third root of unity)
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 +
and since <math>1 + \omega + \omega^2 = 0</math>
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the other coefficients vanish when we add up
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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>
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~Ishan
 
==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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