Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 3: Line 3:
 
==Solution==
 
==Solution==
  
−
<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>
+
<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>
 +
 
 +
<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>
  
 
<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>
 
<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>
  
−
<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>
+
<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>
 +
 
 +
<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>
 +
 
 +
<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>
 +
 
 +
<math> \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. </math>
 +
 
 +
<math>\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. </math>
  
−
<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>
+
<math>\text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. </math>
  
−
<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>
+
\text{ Therefore, the final answer is } \boxed{2^{2005}}. $
  
 
==See also==
 
==See also==

Revision as of 18:23, 17 November 2024

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

$f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies.$ (Error compiling LaTeX. Unknown error_msg)

$\text{This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i.$

$\text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n.$

$\text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}.$

$\text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m.$

$\text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}.$

$\text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003.$

$\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}.$

$\text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}.$

\text{ Therefore, the final answer is } \boxed{2^{2005}}. $

See also

2006 Canadian MO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 Followed by
Problem 2
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_Canadian_MO_Problems/Problem_1&oldid=234765"