In the following grid, fill up the numbers from 1 to 9 (without repetition) such that their sum is $1$.
$\dfrac{\square}{\square \square}+\dfrac{\square}{\square \square} +\dfrac{\square}{\square\square}=1$
Solution provided by Mark, another contributor of this blog:
Analysis for Quiz 10: Training For Problem Solving Skills
The questions I want to give something to ponder, something that is not immediately straightforward, and some cool and rational thought will be required to answer the question.
Admittedly, I've touched a modicum bit in this regard at this slideshow (http://masteringolympiadmathematics.blogspot.com/2015/06/slideshow-9-creative-teaching.html), you could, if you want to refer it and answer the following quiz questions.
Question 1: What is the next square number after $(x-6)^4$?
$((x-6)^2)^2+1$
$((x-6+1)^2)^2$
$((x-6)^2+1)^2$
Answer:
Admittedly, I've touched a modicum bit in this regard at this slideshow (http://masteringolympiadmathematics.blogspot.com/2015/06/slideshow-9-creative-teaching.html), you could, if you want to refer it and answer the following quiz questions.
Question 1: What is the next square number after $(x-6)^4$?
$((x-6)^2)^2+1$
$((x-6+1)^2)^2$
$((x-6)^2+1)^2$
Answer:
Optimization Contest Problem: Find the maximum of f(x) (Heuristic Solution)
Find the maximum of $f(x) = \dfrac{x^4-x^2}{x^6+2x^3-1}$ where $x>1$.
If we want to maximize $f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$, this could be done if we are to find the minimum value for the expression $\dfrac{x^6+2x^3-1}{x^4-x^2}$.
Note that
Subscribe to:
Posts (Atom)