2021 JMPSC Invitationals Problems/Problem 5: Difference between revisions
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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 | 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 | ||
==Solution 2== | |||
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> | |||
~Geometry285 | |||
==See also== | ==See also== | ||
Revision as of 20:09, 11 July 2021
Problem
An 
-pointed fork is a figure that consists of two parts: a handle that weighs 
ounces and "skewers" that each weigh a nonzero integer weight (in ounces). Suppose is a positive integer such that there exists an -pointed fork with weight 
What is the sum of all possible values of ?
Solution
If each skewer weights 
ounces, where must be a positive integer, then the total weight of our fork is 
We equate this to 
and rearrange to get ![\[12+an=n^2\]](http://latex.artofproblemsolving.com/d/3/d/d3d5cf92a41b35c47a28e5d0fcadb9cd579a9c68.png)
![\[an=n^2-12\]](http://latex.artofproblemsolving.com/c/7/c/c7c9621c584762ce732aa9d4562626af8f43e4a5.png)
![\[a=n-\frac{12}{n}.\]](http://latex.artofproblemsolving.com/b/b/9/bb91ce24e6bb47d7433298836d1681ba699543bd.png)
If is an integer and 
is not, it is clear that will not be an integer. Thus, since is an integer, the only possible values of that yield an integer are factors of : ![\[n=1,2,3,4,6,12.\]](http://latex.artofproblemsolving.com/0/b/6/0b6b18612cb23307e22447a5a1e2218adae8445e.png)
Note that is negative for 
and so the only valid are 
leading to an answer of 
. ~samrocksnature
Solution 2
Suppose the integer weight is 
: we have 
. Now, we have 
, so we can have 
, 
, and 
to ensure is positive. Therefore, 
~Geometry285
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
