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

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The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term <math>n</math> (the number of grapes eaten by the child in <math>1</math>-st place), difference <math>d=-2</math>, and number of terms <math>c</math>. We can easily compute that this sum is equal to <math>c(n-c+1)</math>.
 
The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term <math>n</math> (the number of grapes eaten by the child in <math>1</math>-st place), difference <math>d=-2</math>, and number of terms <math>c</math>. We can easily compute that this sum is equal to <math>c(n-c+1)</math>.
  
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Hence we have the equation <math>2009=c(n-c+1)</math>, and we are looking for a solution <math>(n,c)</math>, where both <math>n</math> and <math>c</math> are positive integers, <math>n\geq 2(c-1)</math>, and <math>n</math> is maximized. (The condition <math>n\geq 2(c-1)</math> states that even the last child had to eat a non-negative number of grapes.)
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Hence we have the equation <math>2009=c(n-c+1)</math>, and we are looking for a solution <math>(n,c)</math>, where both <math>n</math> and <math>c</math> are positive integers, <math>n\geq 2(c-1)</math>, and <math>n</math> is minimized. (The condition <math>n\geq 2(c-1)</math> states that even the last child had to eat a non-negative number of grapes.)
  
 
The prime factorization of <math>2009</math> is <math>2009=7^2 \cdot 41</math>. Hence there are <math>6</math> ways how to factor <math>2009</math> into two positive terms <math>c</math> and <math>n-c+1</math>:
 
The prime factorization of <math>2009</math> is <math>2009=7^2 \cdot 41</math>. Hence there are <math>6</math> ways how to factor <math>2009</math> into two positive terms <math>c</math> and <math>n-c+1</math>:
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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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=== Solution 2 ===
  
 
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>.
 
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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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>.
 
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>.
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=== Solution 3 (similar to solution 1) ===
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If the winner ate n grapes, then 2nd place ate <math>n+2-4=n-2</math> grapes, 3rd place ate <math>n+2-6=n-4</math> grapes, 4th place ate <math>n-6</math> grapes, and so on. Our sum can be written as <math>n+(n-2)+(n-4)+(n-6)\dots</math>. If there are x places, we can express this sum as <math>(x+1)n-x(x+1)</math>, as there are <math>(x+1)</math> occurrences of n, and <math>(2+4+6+\dots)</math> is equal to <math>x(x+1)</math>. This can be factored as <math>(x+1)(n-x)=2009</math>. Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If <math>x+1=41</math>, then <math>x=40</math> and <math>n=40+49=\boxed{089}</math>. (Note we would have come upon the same result had we used <math>x+1=49</math>.)
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~MC413551
  
 
== 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}}
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 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:40, 12 January 2024

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 $\boxed{089}$.

Solution 3 (similar to solution 1)

If the winner ate n grapes, then 2nd place ate $n+2-4=n-2$ grapes, 3rd place ate $n+2-6=n-4$ grapes, 4th place ate $n-6$ grapes, and so on. Our sum can be written as $n+(n-2)+(n-4)+(n-6)\dots$. If there are x places, we can express this sum as $(x+1)n-x(x+1)$, as there are $(x+1)$ occurrences of n, and $(2+4+6+\dots)$ is equal to $x(x+1)$. This can be factored as $(x+1)(n-x)=2009$. Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If $x+1=41$, then $x=40$ and $n=40+49=\boxed{089}$. (Note we would have come upon the same result had we used $x+1=49$.) ~MC413551

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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