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

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==Problem==
 
==Problem==
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An <math>n</math>-pointed fork is a figure that consists of two parts: a handle that weighs <math>12</math> ounces and <math>n</math> "skewers" that each weigh a nonzero integer weight (in ounces). Suppose <math>n</math> is a positive integer such that there exists a fork with weight <math>n^2.</math>  What is the sum of all possible values of <math>n</math>?
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An <math>n</math>-pointed fork is a figure that consists of two parts: a handle that weighs <math>12</math> ounces and <math>n</math> "skewers" that each weigh a nonzero integer weight (in ounces). Suppose <math>n</math> is a positive integer such that there exists an <math>n</math>-pointed fork with weight <math>n^2.</math>  What is the sum of all possible values of <math>n</math>?
 
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[[File:Invites5.png|200px]]
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==Solution==
 
==Solution==
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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
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==Solution 2==
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Suppose the integer weight is <math>k</math>: we have <math>n^2-nk-12=0</math>. Now, we have <math>12=2^2 \cdot 3</math>, so we can have <math>(n-12)(n+1)</math>, <math>(n-6)(n+2)</math>, and <math>(n-4)(n+3)</math> to ensure <math>k</math> is positive. Therefore, <math>n=\{4,6,12 \} \implies 4+6+12=\boxed{22}</math>
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~Geometry285
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== Solution 3 ==
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Suppose the weight of each "skewer" is <math>k</math> ounces. We then have that the total weight of the fork is <math>12 + nk</math>. This must equal <math>n^2</math>, so <math>12 + nk = n^2 \implies 12 = n^2 - nk = 12 \implies n(n-k) = 12</math>. This means that <math>n</math> must be a factor of <math>12</math>. Also, the total weight must be greater than <math>12</math> ounces, so we have that <math>n^2 > 12</math>. The factors of <math>12</math> that have a square greater than <math>12</math> are <math>4</math>, <math>6</math> and <math>12</math>, so the answer is <math>4+6+12=\boxed{22}</math>.
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~Mathdreams
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==See also==
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#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
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#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 13:00, 12 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 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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