2021 JMPSC Invitationals Problems/Problem 5

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 an $n$-pointed fork with weight $n^2.$ What is the sum of all possible values of $n$?

Invites5.png

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

Solution 2

Suppose the integer weight is $k$: we have $n^2-nk-12=0$. Now, we have $12=2^2 \cdot 3$, so we can have $(n-12)(n+1)$, $(n-6)(n+2)$, and $(n-4)(n+3)$ to ensure $k$ is positive. Therefore, $n=\{4,6,12 \} \implies 4+6+12=\boxed{22}$

~Geometry285

Solution 3

Suppose the weight of each "skewer" is $k$ ounces. We then have that the total weight of the fork is $12 + nk$. This must equal $n^2$, so $12 + nk = n^2 \implies 12 = n^2 - nk = 12 \implies n(n-k) = 12$. This means that $n$ must be a factor of $12$. Also, the total weight must be greater than $12$ ounces, so we have that $n^2 > 12$. The factors of $12$ that have a square greater than $12$ are $4$, $6$ and $12$, so the answer is $4+6+12=\boxed{22}$.

~Mathdreams

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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