Mock IMO contest problem: Solve $8 x^4-3 x^3-464 x^2-850 x+384=0$.

Mock IMO contest problem:

Solve $8 x^4-3 x^3-464 x^2-850 x+384=0$.

Observe! The best problem solver is also the best observer, train your eyes to see things others won't necessarily see or on first glance. The eyes are the windows to everything.

We need to exercise our mind by solving intriguing challenges as they can certainly keep our  brain sharp. Sharp mindedness means the quickness of a person to interpret, understand, analyze,  or pick up an idea adequately. It is a quality that enables you to find a solution to a problem very quickly or answer a question very quickly or learn very quickly. A person with sharp brain is a keen and quick observer.

China IMO Mock Problem: Given $xy$ is rational number, and $x$ and $y$ are the roots of the equations $6x^2+2015x+8=0$ and $6x^2+2015x+8=0$ respectively. Evaluate $\dfrac{x}{y}$.

Given $xy\ne 1$, and $x$ and $y$ are the roots of the equations $6x^2+2015x+8=0$ and $6x^2+2015x+8=0$ respectively. Evaluate $\dfrac{x}{y}$.

Solution:

Since the coefficients of $6x^2+2015x+8=0$ are the same as the coefficients of $8y^2+2015y+6$ in reversed order, and we're told that $x$ is the root of the equation $6x^2+2015x+8=0$ while $y$ is the root of the equation  $8y^2+2015y+6$.

Slideshow 8: Creating Curious Students/Thinkers

Edward Witten, the World's Greatest Living Theoretical Physicist

 Source from Wikipedia (Edward_Witten):

Witten has been honored with numerous awards including a MacArthur Grant (1982), the Fields Medal (1990), the Nemmers Prize in Mathematics (2000), the National Medal of Science[18] (2002), Pythagoras Award[19] (2005), the Henri Poincaré Prize (2006), the Crafoord Prize (2008), the Lorentz Medal (2010) the Isaac Newton Medal (2010) and the Fundamental Physics Prize (2012). Since 1999, he has been a Foreign Member of the Royal Society (London).[20] Pope Benedict XVI appointed Witten as a member of the Pontifical Academy of Sciences (2006). He also appeared in the list of TIME magazine's 100 most influential people of 2004. In 2012 he became a fellow of the American Mathematical Society.

November 13, 2008 From Discover Magazine 

Quiz 8: Edward Witten, the World's Greatest Living Theoretical Physicist

Solve $\sqrt{x+16}+\sqrt[4]{x+16}=12$.

Solve $\sqrt{x+16}+\sqrt[4]{x+16}=12$.

For some students, they would think to square the given equation three times to get rid of the fourth root and square root terms:

$\sqrt{x+16}+\sqrt[4]{x+16}=12$

$(\sqrt[4]{x+16})^2=(12-\sqrt{x+16})^2$

$\sqrt{x+16}=144-24\sqrt{x+16}+x+16$

Other Heuristic Method to Prove the Given Expression Greater Than Zero

Given that $x\ne 0$ and $x$ is real that satisfies $\large \sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.

Regarding to the previous blog post(china-imo-contest-problem) that I mentioned how to prove

$a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x> 0$ in

China IMO Contest Problem: Given $\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.

Given that $x\ne 0$ and $x$ is real that satisfies $\large \sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.

Answer:

Let $\large a=\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$.

Consider  $\large a=\sqrt[3]{x^5-20x}$, we have:

$\large a^3=x^5-20x---(1)$

Consider $\large a=\sqrt[5]{x^3+20x}$, we have:

You're the mean one!

IMO Problem: If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

We all know in solving this problem correctly, our method should focus on turning the given equation so that the LHS of the equation could be represented by the wanted expression. That is, we hope to turn $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$ so the LHS can take the form $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

2015 IMO contest problem: If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

There is something that this Olympiad problem might trick us because it's obvious that $x^2,\,4y^2,\,9z^2$ are squares of $x,\,2y,\,3z$ and hence, one has reason to believe that the proper first step in solving this problem is to square the first given equation:

$x+2y+3z=6$

Putnam Math Exam Problem: Evaluate $\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}$

Putnam Math Exam Problem:

Evaluate $\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}$.

I admit it, the first (or should I say, the only) thing that we could do about the given fraction's expression, is to factor it:

Second Solution to evaluate $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$

Express $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$ in the form $a+b\sqrt{c}$, where $a,\,b,\,c$ are positive integers.

Answer:

Other solution:

It's always better to rewrite the given expression so both numerator and denominator carry more terms for better visualization effect:

Simplify $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$

Express $\dfrac{\cos 1^{\circ}+\cos 2^{\circ}+\cdots+\cos 44^{\circ}}{\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 44^{\circ}}$ in the form $a+b\sqrt{c}$, where $a,\,b,\,c$ are positive integers.

(This question showed up here(math-teacher-guide) but I did not include its solution in that slide show).

My solution:

Normally for simplifying trigonometric question such as this one, there is one thing that's so worth noticing:

Slideshow 7: One Criterion of Mathematically Proficient Students

Method 2: Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$. Find $x,\,y$.

Method 2:

Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$.

Find $x,\,y$.

IMO Practice Problem: Find the exact real root for the equation $10x^3-12x^2-6x-1=0$

Find the exact real root for the equation $10x^3-12x^2-6x-1=0$.

The given cubic cannot be factored easily and beautifully, so the method of factoring the given polynomial is out of the question.

I hear you, the next best approach might be to try out the substitution method, with the hope that after the substitution, we have less variable terms in our newly set equation. But there seems no suitable substitution is available so to make simpler the given equation.

Slideshow 6: Real World Application of Mathematics

When drawing the graph...

Third Heuristic Method In Solving Problem: Compare $M$ and $N$.

Given that $p,\,q,\,r,\,s,\,a,\,b$ are positive real numbers with $M=\sqrt{ap+br}\cdot \sqrt{\dfrac{q}{a}+\dfrac{s}{b}}$ and $N=\sqrt{pq}+\sqrt{rs}$. Compare $M$ and $N$.

The third inspiring methods of solving the above question will be discussed in this post.

It's solved by one Indian genius, a math friend of mine who compared the quantities between $M^4$ and $N^4$:

He first noticed

Second Heuristic Method In Solving Problem: Compare $M$ and $N$.

Given that $p,\,q,\,r,\,s,\,a,\,b$ are positive real numbers with $M=\sqrt{ap+br}\cdot \sqrt{\dfrac{q}{a}+\dfrac{s}{b}}$ and $N=\sqrt{pq}+\sqrt{rs}$. Compare $M$ and $N$.

One of the other two inspiring methods of solving the above question will be discussed in this post.

When we're asked to compare two quantities, like in this instance, we have $M$ and $N$. But that doesn't restrict us in working and considering only with the linear power of  $M$ and $N$.

Compare M and N

Given that $p,\,q,\,r,\,s,\,a,\,b$ are positive real numbers with $M=\sqrt{ap+br}\cdot \sqrt{\dfrac{q}{a}+\dfrac{s}{b}}$ and $N=\sqrt{pq}+\sqrt{rs}$. Compare $M$ and $N$.

This question has the best form of quality because it allows us to approach it in at least 3 different ways, 3 of which that are enlighting and inspiring.

Mock Olympiad Math Problem: Solve $\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$.

Solve $\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$.

$\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}\left(1-\dfrac{3^{\cos^4 x-\sin^2 x}}{3^{\sin^4 x-\cos^2 x}}\right)=\cos 2x$

Another method to prove $7 ≥ \sqrt 2+\sqrt 5 + \sqrt {11}$

Show with proof which of these two values is smaller:

$7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$

In my (few) previous blog post (Which is greater), I mentioned of how I proved for $\sqrt 2+\sqrt 5 + \sqrt {11}$ is smaller than $7$.

But that doesn't mean that solution is the only way out to prove for that kind of problem.

Analysis Quiz 7: IMO Mock Trigonometric Math Quiz

Analysis Quiz 6:

IMO Mock Trigonometric Math Quiz

Please answer the following questions based on the trigonometric equation below:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

Question 1.

Would you convert the cosecant function to the sine function to solve the trigonometric equation above?

Yes.
No.
Perhaps.

Answer:

For an old hand like me (I'm not old at all, hehehe...), I could say right up front, loudly that I would not convert the cosecant function to the sine function in this case! But, if you are new and eager to learn, I will lead you to the answer, just that you have to be patient.

Quiz 7: IMO Mock Trigonometric Math Quiz

A hilarious math joke: 2+2=4

This picture is taken from hearing-comicsjokes

Show with proof which of these two values is smaller: $7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$

Show with proof which of these two values is smaller:

$7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$

I know this problem could be solved using numerous different methods but I am going to post my solution anyway, since I think and I believe I have done a great job in proving it using the most elementary method and the best of observation and thus it earned a place at this blog. :D

One Worked Example of Solving Heuristic IMO Math Problem

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.

The solution from the U.K. mathematician:

IMO problem that could be solved easily by the linear homogeneous recursion method

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.

Terence Tao: The World Greatest Mathematician

Some excerpts about the biography and achievement of Terence Tao, or Terry, as he's mostly known, from the major newspapers from the past 9 years:

March 7, 2015

Terence Tao: the Mozart of maths from The Sydney Morning Herald

When he was nine, Tao commenced part-time studies in mathematics at Flinders University. By the time he was 16, he'd finished his science degree. He got his masters when he was 17 and his PhD at Princeton University at 20.

"He is arguably the world's best mathematician," says Joseph Rudnick, the dean of Physical Sciences at the University of California, Los Angeles (UCLA), where Tao, now 39, has been a mathematics professor since he was 24. "Other mathematicians speak of him in tones of awe." His talents, says Rudnick, are "other-worldly".

Tao has pages of awards, fellowships, prizes and medals to his name - most notably, the Fields Medal, the maths world's equivalent of the Nobel Prize, which he received in 2006 when he was 31 "for his contributions to partial differential equations, combinatorics, harmonic analysis and additive number theory". Around about then, people started to describe him as "the Mozart of math".

Mathematical Problem Solving Skill

In this blog post, I will show one really insightful method given by one very intelligent U.K. mathematician on how he used his own elegant way to prove that $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$.

He first noticed that

$\tan 3(20^\circ)=\tan 60^\circ=\sqrt{3}$

$\tan 3(40^\circ)=\tan 120^\circ=-\sqrt{3}$

$\tan 3(80^\circ)=\tan 240^\circ=\sqrt{3}$

Olympiad Algebra Problem: Evaluate $ax^5+by^5$

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.

Third Method in Proving $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$

As a good mathematics educator, we should solve the best quality math problems using as many ways as we could, because each solution has its own learning value. Who knows, students might gain something from the long and tedious method of problem solving method and one day, they will surprise us with the improved version of the solution!

One thing every mathematics educator has to remember is that our imaginations are vibrant, our hearts are open, everything about math amazes us, and we think anything is possible.

Alternative Way to Prove $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

So in one of my recent post, I mentioned of one rare but extremely useful trigonometry identity that sounds:

$\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

We should then use it when appropriate.

Quiz of World Most Famous Mathematician: Terence Tao

Challenging Math Contest Problem: Prove that $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$

Given $\tan x+\tan (x+60^{\circ})-\tan (60^{\circ}-x)=3\tan 3x$,

prove that $\tan^2 x+\tan^2 (x+60^{\circ})+\tan^2 (60^{\circ}-x)=9\tan^2 3x+6$.

Wow! This is another exquisite problem that one cannot afford to pass it up but to take it as one tough learning example problem so to train students to be the best.  Remember that good teachers will forever encourage learning for understanding and are concerned with developing their students’ critical-thinking skills, problem-solving skills, and problem-approach behaviors. This problem fulfills the these goals of training students to think creatively and that is the reason I bring it to our table.

I won't beat about the bush, so here goes my plan of attack:

Olympiad Trigonometry Problem

Prove $\tan^2 20^\circ+\tan^2 40^\circ+\tan^2 80^\circ=33$.

Good trigonometry problem is hard to come by, and when we, the math educator found one, we have to take advantage of it and make full use of it.

Of course, we can rest assured that this trigonometry problem can be tackled using the sum-to-product and product-to-sum identities in a really messy and tedious way. After all, any given math problem can be solved in the most traditional way, isn't it?

Exploration of some real life applications of mathematics.

It's high time to launch the campaign to increase awareness of how useful math can be to us, how math helps us in our daily life activities, by us, yes, I mean for just anyone else, a student, a worker, a boss, or a housewife.

But does that mean government has to fork out a handsome sum of money to do that? Not really, in fact, if we could bring in the real scenarios where and when math could be of help in our life, then that alone helps to entice the even most reticent students who always find excuses to hate math, as according to them, math is too big for their brain. If we could show to students some really amazing true stories, that could kindle their passion for math and this doesn't need much money and this has everything to do with if we are capable enough to think of attracting real scenarios and bring them to the classroom.

Slideshow 5: Recruit For Attitude, Train For Skills

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Quiz 5: Training For Heuristic Problem Solving Skills

Math Olympiad Problem: Solve for real solutions

Solve for real solutions for the equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$.

Okay, I heard you, why on earth this problem is supposed to be a delicious question that can promote higher mathematics thinking skills?

One of the most common issues math educators are struggling with is the students who underestimate the so-called trivial math problem and they think by the long and typical tedious solving method, the trivial math problem could be safely and successfully solved without a hitch.

Seeing Pattern Seeing Relationship--- Problem Solving Skill

In my previous post(Heuristic-math-problem-solving-skill), I asked the readers to find the sum of the first $n$ terms for the series:

$1+2(1-x)+3(1-x)(1-2x)+4(1-x)(1-2x)(1-3x)+\cdots+n(1-x)(1-2x)\cdots(1-(n-1)x)$

This is one really hard math Olympiad contest problem, but fortunately logical thinking and heuristic problem solving skills are something that we can teach.

Heuristic Math Problem Solving Skill: Seeing Pattern

Brilliant tudents know how and when to implement heuristic strategies. Heuristic strategies are devices which drastically limits the search for a solution in a large problem space (Polya, 2004).

They provide a general suggestion or technique for solving different types of problems and can be used independently or in combination. The ten most common strategies include:

Working backwards, Finding a pattern, Adopting a different point of view, Solving a simpler analogous problem, Considering extreme cases, Making a drawing, Intelligent guessing and testing, Accounting for all possibilities, Organizing data and logical reasoning (Posamentier, Smith and Stepelman, 2006).

Intermediate Value Theorem to Prove Inequality Problem

In my previous post, I talked about how to use calculus method to prove that $\sin 10^{\circ} > \dfrac{1}{6}$ (Never Be Complacent).

Or more precisely, the Intermediate Value Theorem as our weapon to prove it.

We will use the triple angle identity for $\sin 3x=3\sin x-4\sin^3 x$ and the value for $\sin 30^{\circ}=\dfrac{1}{2}$ when $x=10^{\circ}$ in our solution:

Never be complacent of your math problem solving skills

If you want to become a legendary successful learner/student, there is one golden rule to abide by:

Never become complacent with what you have accomplished in the past. Always persuade and inspire your inner self to explore new ideas and venture into new territories so to develop 21st century competencies and become a critical thinker and problem solver.

Small Angle Approximation in Trigonometry

Our previous blog post used the method of proving by contradiction to prove for the inequality $\sin 10^{\circ}>\dfrac{1}{6}$ (Prove that $\sin 10^{\circ}>\dfrac{1}{6}$).

Alternatively, we could also use the "Small-angle Approximation" to prove that $\sin 10^{\circ}>\dfrac{1}{6}$, provided we know fully well that as the angle approaches zero (or the angle is less than $0.2 \,\,\text{radians}$, it is clear that the gap between the approximation and the original function quickly vanishes.
 (Graph taken from Small-angle_approximation (Wikipedia))

Prove that $\sin 10^{\circ}>\dfrac{1}{6}$.

Another heuristic problem solving technique that is so popular and widely use is called "Prove by Contradiction" technique.

Euclid's beautiful proof by contradiction to show there are infinitely many prime numbers is one of the most mind-boggling method to prove something is true that leave us speechless with absolute admiration.

Show that $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

Show that $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

On my previous post (Show that $\displaystyle 16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17$), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}$. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}$ will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

Prove 16<1/√(1)+1/√(2)+1/√(3)+1/√(4)+...1/√(79)+1/√(80)< 17

Show that $\displaystyle 16 < \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}< 17$.

It goes without saying that there's more than one way to skin any given math problem. Exposing students with intriguing math problem can actually instill a genuine interest in creative problem-solving from an early age.

Heuristic Method for Solving Hard System

Solve the system

$\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}$

$\dfrac{1}{y}+\dfrac{1}{x+z}=\dfrac{1}{3}$

$\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}$

Method II (Heuristic and satisfactory solution):

Solving Hard System

Solve the system

$\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}$

$\dfrac{1}{y}+\dfrac{1}{x+z}=\dfrac{1}{3}$

$\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}$

Method I (Long and tedious but workable solution):

Evaluate ab+bc+ca

Given real numbers $a,\,b,\,c$ such that

$a^2+ab+b^2=2$, $b^2+bc+c^2=3$, $c^2+ca+a^2=5$

Evaluate $ab+bc+ca$.

If nothing that we could do to algebraically manipulate this problem that comes to mind, as this is a hard problem, then what is left for us to be done, is, we can sum the three given equations, to get:

A Challenge Problem of Simultaneous Equation

Given real numbers $a,\,b,\,c$ such that

$a^2+ab+b^2=2$, $b^2+bc+c^2=3$, $c^2+ca+a^2=5$

Evaluate $ab+bc+ca$.

I hope you would try this problem out and attempt it in every way you could. Break this problem down into smaller cases and solve each piece, experience the process and conquer it!

I will post the suggested solution soon. See you!

Slideshow 4: Mathematics Quizzes