2009 AIME II Problems/Problem 4

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

Let us go in order of place, from $1$ to $k$ . The first kid eats $n$ grapes, the second kid eats $n - 2$ grapes, and so on. These must sum up to 2009. Hence $n(k + 1) - 2(k)(k + 1)/2 = 2009$ , or $n(k + 1) - k(k + 1) = 2009$ . Factoring, we get $(n - k)(k + 1) = 2009$ . We are only looking for integer solutions, so we can prime factorize 2009 and look for factor pairs. $2009 = (7)*(7)*(41)$ . So $(n - k) = 2009$ and $k + 1 = 1$ , $(n - k) = 49$ and $(k + 1) = 41$ , or $(n - k) = 287 and$ (k + 1) = 7 $. The solution sets (in the form (n , k), respectively, are$ (2009, 0) $,$ (293, 6) $, and$ (89, 48) $. The pair with the smallest n value has$ n=\boxed{089}$.

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
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