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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 5"

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If each skewer weights <math>a</math> ounces, where <math>a</math> must be a positive integer, then the total weight of our fork is <math>12+an.</math> We equate this to <math>n^2</math> and rearrange to get <cmath>12+an=n^2</cmath> <cmath>an=n^2-12</cmath> <cmath>a=n-\frac{12}{n}.</cmath> If <math>n</math> is an integer and <math>\frac{12}{n}</math> is not, it is clear that <math>a</math> will not be an integer. Thus, since <math>n</math> is an integer, the only possible values of <math>n</math> that yield an integer <math>a</math> are factors of <math>12</math>: <cmath>n=1,2,3,4,6,12.</cmath> Note that <math>a</math> is negative for <math>n=1,2,3</math> and so the only valid <math>n</math> are <math>4,6,12,</math> leading to an answer of <math>4+6+12=\boxed{22}</math>. ~samrocksnature

Revision as of 15:45, 11 July 2021

Problem

An $n$-pointed fork is a figure that consists of two parts: a handle that weighs $12$ ounces and $n$ "skewers" that each weigh a nonzero integer weight (in ounces). Suppose $n$ is a positive integer such that there exists a fork with weight $n^2.$ What is the sum of all possible values of $n$?

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Solution

If each skewer weights $a$ ounces, where $a$ must be a positive integer, then the total weight of our fork is $12+an.$ We equate this to $n^2$ and rearrange to get \[12+an=n^2\] \[an=n^2-12\] \[a=n-\frac{12}{n}.\] If $n$ is an integer and $\frac{12}{n}$ is not, it is clear that $a$ will not be an integer. Thus, since $n$ is an integer, the only possible values of $n$ that yield an integer $a$ are factors of $12$: \[n=1,2,3,4,6,12.\] Note that $a$ is negative for $n=1,2,3$ and so the only valid $n$ are $4,6,12,$ leading to an answer of $4+6+12=\boxed{22}$. ~samrocksnature

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