Difference between revisions of "1991 OIM Problems/Problem 5"
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== Solution == | == Solution == | ||
− | * Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. | + | |
− | {{ | + | '''Part i.''' |
+ | |||
+ | Let <math>x</math>, <math>y</math>, <math>P</math> be integers | ||
+ | |||
+ | <math>2x^2 - 6xy + 5y^2-P=0</math>, then solving for <math>x</math> using the quadratic equation we have: | ||
+ | |||
+ | <math>x=\frac{3y \pm \sqrt{2P-y^2}}{2}</math> | ||
+ | |||
+ | Let <math>K</math> be an integer and <math>K^2=2P-y^2</math>. Since <math>1 \le P \le 100</math>, then | ||
+ | |||
+ | * Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one. | ||
+ | |||
+ | {{Alternate solutions}} | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe6.htm | https://www.oma.org.ar/enunciados/ibe6.htm |
Revision as of 19:59, 22 December 2023
Problem
Let . We will say that an integer is a value of if there exist integers and such that .
i. Determine how many elements of {1, 2, 3, ... ,100} are values of .
ii. Prove that the product of values of is a value of .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Part i.
Let , , be integers
, then solving for using the quadratic equation we have:
Let be an integer and . Since , then
- Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.