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Difference between revisions of "2009 AIME II Problems/Problem 4"

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The smallest valid solution is therefore <math>c=41</math>, <math>n=\boxed{089}</math>.
 
The smallest valid solution is therefore <math>c=41</math>, <math>n=\boxed{089}</math>.
  
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=== Another Solution ===
  
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=== Another Solution ===
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If the first child ate <math>n=2m</math> grapes, then the maximum number of grapes eaten by all the children together is <math>2m + (2m-2) + (2m-4) + \cdots + 4 + 2 = m(m+1)</math>. Similarly, if the first child ate <math>2m-1</math> grapes, the maximum total number of grapes eaten is <math>(2m-1)+(2m-3)+\cdots+3+1 = m^2</math>.
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For <math>m=44</math> the value <math>m(m+1)=44\cdot 45 =1980</math> is less than <math>2009</math>. Hence <math>n</math> must be at least <math>2\cdot 44+1=89</math>. For <math>n=89</math>, the maximum possible sum is <math>45^2=2025</math>. And we can easily see that <math>2009 = 2025 - 16 = 2025 - (1+3+5+7)</math>, hence <math>2009</math> grapes can indeed be achieved for <math>n=89</math> by dropping the last four children.
  
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Let us go in order of place, from <math>1</math> to <math>k</math>. The first kid eats <math>n</math> grapes, the second kid eats <math>n - 2</math> grapes, and so on. These must sum up to 2009. Hence <math>n(k + 1) - 2(k)(k + 1)/2 = 2009</math>, or <math>n(k + 1) - k(k + 1) = 2009</math>. Factoring, we get <math>(n - k)(k + 1) = 2009</math>. We are only looking for integer solutions, so we can prime factorize 2009 and look for factor pairs. <math>2009 = (7)*(7)*(41)</math>. So <math>(n - k) = 2009</math> and <math>k + 1 = 1</math>, <math>(n - k) = 49</math> and <math>(k + 1) = 41</math>, or <math>(n - k) = 287</math> and <math>(k + 1) = 7</math>. The solution sets (in the form (n , k), respectively, are <math>(2009, 0)</math>, <math>(293, 6)</math>, and <math>(89, 48)</math>. The pair with the smallest n value has <math>n=\boxed{089}</math>.
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Hence we found a solution with <math>n=89</math> and <math>45-4=41</math> kids, and we also showed that no smaller solution exists. Therefore the answer is <math>\boxed{089}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2009|n=II|num-b=3|num-a=5}}

Revision as of 12:12, 9 April 2009

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 $n$.

Solving this problem

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 maximized. (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 $c=41$, $n=\boxed{089}$.

Another Solution

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
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