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 "2009 AIME II Problems/Problem 4"

m (→‎Solution 1)
Line 38: Line 38:
  
 
{{AIME box|year=2009|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2009|n=II|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Revision as of 22:36, 4 July 2013

Problem

A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$-th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$. Find the smallest possible value of $n$.

Solution

Resolving the ambiguity

The problem statement is confusing, as it can be interpreted in two ways: Either as "there is a $k>1$ such that the child in $k$-th place had eaten $n+2-2k$ grapes", or "for all $k$, the child in $k$-th place had eaten $n+2-2k$ grapes".

The second meaning was apparrently the intended one. Hence we will restate the problem statement in this way:

A group of $c$ children held a grape-eating contest. When the contest was over, the following was true: There was a $n$ such that for each $k$ between $1$ and $c$, inclusive, the child in $k$-th place had eaten exactly $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$. Find the smallest possible value of $n$.

Solution 1

The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$-st place), difference $d=-2$, and number of terms $c$. We can easily compute that this sum is equal to $c(n-c+1)$.

Hence we have the equation $2009=c(n-c+1)$, and we are looking for a solution $(n,c)$, where both $n$ and $c$ are positive integers, $n\geq 2(c-1)$, and $n$ is minimized. (The condition $n\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.)

The prime factorization of $2009$ is $2009=7^2 \cdot 41$. Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$:

  • $c=1$, $n-c+1=2009$, then $n=2009$ is a solution.
  • $c=7$, $n-c+1=7\cdot 41=287$, then $n=293$ is a solution.
  • $c=41$, $n-c+1=7\cdot 7=49$, then $n=89$ is a solution.
  • In each of the other three cases, $n$ will obviously be less than $2(c-1)$, hence these are not valid solutions.

The smallest valid solution is therefore $c=41$, $n=\boxed{089}$.

Solution 2

If the first child ate $n=2m$ grapes, then the maximum number of grapes eaten by all the children together is $2m + (2m-2) + (2m-4) + \cdots + 4 + 2 = m(m+1)$. Similarly, if the first child ate $2m-1$ grapes, the maximum total number of grapes eaten is $(2m-1)+(2m-3)+\cdots+3+1 = m^2$.

For $m=44$ the value $m(m+1)=44\cdot 45 =1980$ is less than $2009$. Hence $n$ must be at least $2\cdot 44+1=89$. For $n=89$, the maximum possible sum is $45^2=2025$. And we can easily see that $2009 = 2025 - 16 = 2025 - (1+3+5+7)$, hence $2009$ grapes can indeed be achieved for $n=89$ by dropping the last four children.

Hence we found a solution with $n=89$ and $45-4=41$ kids, and we also showed that no smaller solution exists. Therefore the answer is $\boxed{089}$.

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_4&oldid=55469"