Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 5"
Geometry285 (talk | contribs) m |
|||
Line 8: | Line 8: | ||
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
Contents
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 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 : 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.