With a rush of adrenaline, I made the daring decision to clone the repository and build that game of *torture*. And it worked. It’s called `./opensuplaplex`

now.

^{This is me beating the 3rd level.}

This is a “follow-up” to the previous article: “The math exams of my life”, as some readers were curious to see some examples of Math Olympiad exercises.

This a selection of *cute*, *non-trivial* algebra problems (with a hint of *number theory*) *compiled* from the Romanian Math Olympiad (regional phase or *faza judeteana*) for 8^{th}, 9^{th}, and 10^{th} graders (13-15 years old).

The solutions are surprising and involve a good understanding of algebraic concepts, pattern spotting, or tricks that, in the long run, help students develop mathematical intuition.

Depending on your passion for mathematics (or competitive mathematics), the problems should pose enough difficulty to keep you entertained for a few hours. If you are stuck with one problem, try to read the hint instead of going straight to the answer.

In case you want to solve them by yourself, do a short recap on the following subjects:

- Sets;
- Sequences of numbers;
- Faulhaber’s formula;
- Rearrangement inequality
- AM-GM Inequality;
- Cauchy–Bunyakovsky–Schwarz Inequality;
- Hermite’s Identity;
- Monotonic functions.

The main topic of this problem set is: “Inequalities”.

I have a few notebooks containing solutions for various Math problems I’ve solved over the years (recreational mathematics). If time allows, I will publish more lists covering more topics. Currently, I am in the process of grouping them into categories.

^{“Screenshot” from one my notebooks.}

The problems are from the “Regional” Phase of the Romanian Math Olympiad. The truly difficult problems are usually the ones from the “National” phase. I am planning to publish a list of those as well.

Another important aspect is that I am not a *mathematician*, so if you see that solutions are incorrect or have better solutions, please send me some feedback.

**1.** ^{easy} Compute \(S=1-4+9-16+..+99^2-100^2\). ^{Hint & Answer}

**2.** ^{easy} Determine the smallest element of the set \(\{ab \text{ } \lvert \text{ } a,b \in \mathbb{R} \text{ and } a^2 + 2b^2=1\}\)? ^{Hint & Answer}

**3.** ^{easy} What is the cardinality of following set: \(\{x \in \mathbb{R} \lvert [\frac{x+1}{5}]=\{\frac{x-1}{2}\} \}\). \(\{ a \}\) is the fractional part of the number \(a \in \mathbb{R}\), and \([a]\) is the integer part of \(a \in \mathbb{R}\). ^{Hint & Answer}

**4.** ^{easy} Find all the elements of the set \(\{\frac{3}{2x} \lvert x \in \mathbb{R} \text{ and, } \frac{1}{[x]} + \frac{1}{\{x\}}=2x \}\). \(\{ a \}\) is the fractional part of the number \(a \in \mathbb{R}\), and \([a]\) is the integer part of \(a \in \mathbb{R}\). ^{Hint & Answer}

**5.** ^{easy} Given \(a,b,c \in \mathbb{R}^{*}\), we know that \((a,b,c)\) are part of an arithmetic progression, \((ab, bc, ca)\) are part a geometric progression, and \(a+b+c=ab+bc+ca\). The triplets in the set \(M=\{(a,b,c)\ \lvert a,b,c \in \mathbb{R}^{*}\}\) satisfy all the conditions mentioned before.

Compute \(\sum_{(a,b,c) \in M} (\lvert a \rvert + \lvert b \rvert + \lvert c \rvert)\). ^{Hint & Answer}

**6.** ^{easy} For the following sequence of numbers \((a_n)_{n \ge 1}\), \(a_1=1\) and \(a_2=6\) and \(a_{n+1}=\frac{a_n}{a_{n-1}}\), \(n \ge 2\), compute \(a_{2021}\). ^{Hint & Answer}

**7.** ^{medium} Find all numbers \(k \in \mathbb{Z}\), so that \(a^4+b^4+c^4+d^4+k*abcd \ge 0\), \(\forall a,b,c,d \in \mathbb{R}^{*}\). ^{Hint & Answer}

**8.** ^{medium} If \(x,y,z,t \in \mathbb{R}\), and \((x-3y+6z-t)^2 \ge 2021\) and \(x^2+y^2+z^2+t^2 \le 43\), then what is the value of the expression: \(\lvert x + y + z + t \rvert\) ? ^{Hint & Answer}

**9.** ^{medium} Considering \(x^2 + (a+b+c)x + k(ab+bc+ca) = 0\) where \(a,b,c \in \mathbb{R}_{+}^{*} \text{, and } k \in \mathbb{R}\) prove that \(\forall k \le \frac{3}{4}\) the equation has all its solution in \(\mathbb{R}\). ^{Hint & Answer}

**10.** ^{easy} Prove that \([\frac{x+3}{6}] - [\frac{x+4}{6}] + [\frac{x+5}{6}] = [\frac{x+1}{2}] - [\frac{x+1}{3}]\) is true, \(\forall x \in \mathbb{R}\). ^{Hint & Answer}

**11.** ^{medium} Prove that if \(\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} a_k^2\) then \(\sum_{k=1}^{n} a_k \le n\), with \(a_k \in \mathbb{R}_{+}\). ^{Hint & Answer}

**12.** ^{medium} If \(a^2+b^2+c^2=3\) prove (\(\lvert a \rvert + \lvert b \rvert + \lvert c \rvert - abc) \le 4\), where \(a,b,c \in \mathbb{R}\). ^{Hint & Answer}

**13.** ^{hard} Prove \(\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2} \ge 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\) if \(a,b,c \in \mathbb{R}_{+}^{*}\). ^{Hint & Answer}

**14.** ^{easy} Prove that if \(x \in \mathbb{R}\) and \(x^2+x \in \mathbb{Q}\) and \(x^3+2x \in \mathbb{Q}\), then \(x \in \mathbb{Q} \subset \mathbb{R}\). ^{Hint & Answer}

**15.** ^{medium} Given \(a,b \in \mathbb{R}\), we know \(3^a+13^b=17^a\), and \(5^a+7^b=11^b\). Prove \(a \lt b\). ^{Hint & Answer}

**16.** ^{medium} For \(n \in \mathbb{N}, n \ge 2\), let \(u(n)\) be the biggest prime number \(\le n\) and \(v(n)\) be the smallest prime number \(\gt n\). Prove:

^{Hint & Answer}

**17.** ^{medium} Prove \(\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy} \le \frac{1}{2} (\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})\), \(\forall x,y,z \in \mathbb{R}_{+}^*\). ^{Hint & Answer}

**18.** ^{medium} For \(a,b,c \in (0,1) \subset \mathbb{R}\), \(x,y,z \in (0, \infty) \subset \mathbb{R}\), if:

Prove that:

\[\frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z} \le \frac{3}{4}\]^{Hint & Answer}

**19.** ^{easy} If \(x,y,z \in R_{+}^{*}\), and \(xy=\frac{z-x+1}{y}=\frac{z+1}{2}\), prove that one of the numbers is the arithmetical mean of the other two. ^{Answer}

**20.** ^{medium} If \(a,b,c \in (1, \infty) or a,b,c \in (0,1)\). Prove:

^{Hint & Answer}

Try playing with Faulhaber’s formula.

Find a way to *introduce* \(ab\) in the given equality \(a^2 + 2b^2=1\).

Try to get rid of the fractional part.

Try to do some substitutions based on the fact \(x=[x]+\{x\}\).

Use \(a\) to express \(b\) and \(c\).

Look for any patterns by computing the first few terms of the sequence.

Give *meaningful* values to \(a,b,c,d\) and see what’s happening.

Have you considered AM-GM inequality?

Can you use the Cauchy–Bunyakovsky–Schwarz inequality to solve the problem?

Can you use the Rearrangement inequality to solve the problem?

Can you use Hermite’s identity to solve the problem?

Can you use the Cauchy–Bunyakovsky–Schwarz inequality to solve the problem?

Can you use both CBS and AM-GM inequalities?

Can you prove first \(\frac{x}{y^2}+\frac{y}{x^2} \ge \frac{1}{x} + \frac{1}{y}\) ?

Try expressing \(x\) as a relationship between two rational numbers.

Think in terms of monotonically increasing and monotonically decreasing functions.

How many times a term \(\frac{1}{u(n)*v(n)}\) appears in the sum ?

Can you find a way to use the AM-GM inequality?

Work on the expressions involving logarithms. Consider changing the base of the logarithms to a common one.

This exercise is easy, so it doesn’t deserve a hint.

Consider changing the base of the logarithms to a common one.

We write our sum as:

\[S=(1^2-2^2)+(3^2-4^2)+...+(99^2-100^2)\]There is a formula for the difference of two square numbers:

\[a^2-b^2=(a-b)(a+b)\]Our \(S\) is a sum of differences between subsequent square numbers:

\[S=(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100) \Leftrightarrow\] \[S=-3-7-11-...-199\]If you have a keen eye for observations, you notice the numbers \(3,7,11,...,199\) have this form \(3+k*4\), \(k=0..49\).

\[S=-(3+0*4)-(3+1*4)-(3+2*4)-...-(3+49*4) \Leftrightarrow\]\(S=-3*50-4(0+1+2+...+49) \Leftrightarrow\)https://en.wikipedia.org/wiki/Rearrangement_inequality

The infinite series whose terms are natural numbers, \(1+2+3+...+n\), is divergent. But we know the \(nth\) partial sum of the series to be: \(\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\), so:

\[S=-150-4*\frac{49(49+1)}{2}\]The final answer \(S=-150-4900=-5050\).

If you are familiar with Faulhaber’s formula, you know that:

\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\] \[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\]We can cleverly use the two formulas and reimagine our \(S\) to be:

\[S=\underbrace{(1^2+2^2+3^2+...+100^2)}_{\sum_{k=1}^{100} k^2} - \underbrace{2*(2^2+4^2+6^2+..+100^2)}_{\sum_{k=1}^{50} (2k)^2} \Leftrightarrow\] \[S=\frac{100*101*201}{6}-2*(2^2*1^2+2^2*2^2+2^2*3^2+...+2^2*50^2) \Leftrightarrow\] \[S=\frac{100*101*201}{6}-8*\frac{50*51*101}{6}\]Final answer \(S=-5050\).

The intuition begs us to find a way to link our existing relationship to a term containing \(ab\). In this regard:

\[1=a^2+2b^2=a^2 + (b\sqrt{2})^2 \Leftrightarrow\] \[1=a^2 + \underbrace{2*\sqrt{2}ab}_{\text{we add this}} + (b\sqrt{2})^2 - \underbrace{2*\sqrt{2}ab}_{\text{to remove it later}} \Leftrightarrow\] \[1=(a+b\sqrt{2})^2 - 2\sqrt{2}ab \Leftrightarrow\] \[1+2\sqrt{2}ab=(a+b\sqrt{2})^2\]But we know that:

\[(a+b\sqrt{2})^2 \geq 0 \Rightarrow\] \[1+2\sqrt{2}ab \geq 0 \Rightarrow\] \[ab \geq \frac{-1}{2\sqrt{2}}\]But is it possible for \(ab=\frac{-1}{2\sqrt{2}}\). Yes, \(ab=\frac{-1}{2\sqrt{2}}\) when \((a+b\sqrt{2})^2=0\).

All in all, the smallest element of our set is \(\frac{-1}{2\sqrt{2}}\).

If \(a \in \mathbb{R}\), then:

\[a=[a]+\{a\} \Leftrightarrow\] \[\{a\}=a-[a]\]With this in mind, we want to get rid of *the wild* fractional part \(\{a\}\) from our relationship:

We can safely say that \(\frac{x-1}{2}=k \in \mathbb{Z}\).

We substitute \(x=2k-1\) in the original relation:

\[[\frac{2*k+1+1}{5}] + [\frac{2*k+1-1}{2}] = k \Leftrightarrow\] \[[\frac{2k+2}{5}] + k = k \Leftrightarrow\] \[[\frac{2k+2}{5}] = 0\]Because \([\frac{2k+2}{5}] = 0\), then:

\[0 \leq \frac{2k+2}{5} \lt 1 \Leftrightarrow\] \[-1 \leq k \lt \frac{3}{2}\]There are \(3\) numbers \(k \in \mathbb{Z}\) that satisfy the relationship: \(\{-1,0,1\}\). The final answer is \(3\).

To avoid division by zero in \(\frac{3}{2x}\), we know that \(x \neq 0\).

Note: The

fractional part operation(denoted by \(\{a\}\)) is not distributive over multiplication, meaning \(\{a*b\} \neq \{a\} * \{b\}\), so avoid making any further assumptions about \(\frac{1}{\{x\}}\).

Firstly, \(\frac{1}{[x]} + \frac{1}{\{x\}}=2x\) is equivalent to \(\frac{x}{[x]\{x\}}=2x\), is equivalent to \(x(2[x]\{x\}-1)=0\).

Because \(x \neq 0\), then \(2[x]\{x\}=1\), or \(\{x\}=\frac{1}{2[x]}\).

We substitute \([x]=n\), so that \(\{x\}=\frac{1}{2n}\), where \(n \in \mathbb{Z}\) and \(n \geq 1\).

\(x=[x]+{x}\) becomes \(x=n+\frac{1}{2n}=\frac{2n^2+1}{2n}\).

The term defining the set becomes \([\frac{3}{2x}] \rightarrow [\frac{3}{2*\frac{2n^2+1}{2n}}]=[\frac{3n}{2n^2+1}]\).

For \(n \ge 1\), we know that \(0 \lt \frac{3n}{2n^2+1} \lt 1\). Without calculus, we can easily prove that for \(n \ge 1\), \(f(n)=3n\), and \(g(n)=2n^2+1\) are increasing functions, with \(g(n)\) *increasing faster*, so that \(\frac{f(n)}{g(n)} \lt 1\).

For certain we know that \(0 \lt \frac{3n}{2n^2+1}\).

After we solve \(\frac{3n}{2n^2+1} \lt 1\), we obtain \(n \ge 1\).

For \(n=1\), \([\frac{3}{2x}] \rightarrow 1\), for \(n \gt 1\), \([\frac{3}{2x}] \rightarrow 0\).

So the set \(\{\frac{3}{2x} \lvert x \in \mathbb{R} \text{ and, } \frac{1}{[x]} + \frac{1}{\{x\}}=2x \} = \{0, 1\}\)

\((a,b,c)\) are in an arithmetic progression \(\Rightarrow b=\frac{a+c}{2}\).

\((ab, bc, ac)\) are in a geometric progression \(\Rightarrow (bc)^2 = ab*ac\).

With this in mind, we have the following relationships:

\[\begin{cases} b=\frac{a+c}{2} \\ (bc)^2 = ab*ac \\ a+b+c=ab+bc+ca \end{cases}\]It doesn’t look like it, but we have enough *information* to determine \((a,b,c)\)

But we can also write \(2b=a+\frac{a^2}{b}\).

Putting all together in the last equation:

\[3b=ab+\underbrace{bc}_{a^2}+a\underbrace{c}_{\frac{a^2}{b}} \Leftrightarrow \\ 3b=ab+a^2+\frac{a^3}{b} \Leftrightarrow \\ 3b=a(\underbrace{a+\frac{a^2}{b}}_{2b})+ab \Leftrightarrow \\ 3b=2ab+ab\]But because \(b \in \mathbb{R}^{*}\), and \(3b=3ab\), we can safely assume \(a=1\).

If \(a=1\), the relationship \(2b=a+c\) becomes \(2b=1+\frac{1}{b} \Leftrightarrow 2b^2-b-1=0\).

\(2b^2-b-1=0\) can also be written as \(2b^2+b-(2b+1)=0\) or \(b(2b+1)-(2b+1)=0 \Leftrightarrow (b-1)(2b+1)=0\), so \(b=1\) or \(b=-\frac{1}{2}\).

For \(b=1\), \(c=1\), and for \(b=-\frac{1}{2}\), \(c=-2\).

So our triplets are \((1,1,1)\) or \((1,-\frac{1}{2}, -2)\).

Computing the sum is trivial: \(\sum=\frac{13}{2}\).

The key to solving this exercise is to compute the first terms of the sequence:

\[\begin{cases} a_1=2 \\ a_2=6 \\ a_3=\frac{6}{2}=3 \\ a_4=\frac{3}{6}=\frac{1}{2} \\ a_5=\frac{\frac{1}{2}}{3}=\frac{1}{6} \\ a_6=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{3} \\ a_7=\frac{\frac{1}{3}}{\frac{1}{6}}=2 \\ a_8=\frac{2}{\frac{1}{3}}=6 \\ \text{... and so on} \end{cases}\]We can see that after every 6 terms, the values repeat themselves. Using mathematical induction, we can prove that our sequence is:

\[S(n)= \begin{cases} 2 \text{ ,}\text{ if }n=6*k+1 \\ 6 \text{ ,}\text{ if }n=6*k+2 \\ 3 \text{ ,}\text{ if }n=6*k+3 \\ \frac{1}{2} \text{ ,}\text{ if }n=6*k+4 \\ \frac{1}{6} \text{ ,}\text{ if }n=6*k+5 \\ \frac{1}{3} \text{ ,}\text{ if }n=6*k \end{cases} \text{, where } k \in \mathbb{N}\]Final answer is: \(a_{2021} = \frac{1}{6}\)

When you have problems like this, you need to consider giving specific values to \(a,b,c,d\) that can help you narrow down your search.

For example, we know the inequality should hold regardless of \(a, b, c, d\) (\(\forall\)), so let’s assume the following:

\[a=b=c=d=n\]Then:

\[4*n^4+k*n^4 \ge 0 \Leftrightarrow \\ k \ge -4\]This already tells us that we need to look somewhere in \(k \in [-4, \infty) \cap \mathbb{Z}\).

On my first try, I’ve tried with \(a^2=b^2=m\) and \(c^2=b^2=n\), so I’ve got something like \((n\sqrt{2} - m\sqrt{2})^2 + mn*(k+4) \ge 0\).

Now, we need to find an upper bound for \(k\). For this, we need to apply some tricks to obtain:

\[\text{something positive} + (\text{an upper bound}-k)*(\text{something positive}) \ge 0\]So why don’t we pick:

\[\begin{cases} a=b=c=n \\ d=-n \end{cases}\]Then our relationship becomes:

\[4*n^4 - k*n^4 \ge 0 \Leftrightarrow \\ k \le 4\]So, at this point, we know that \(k \in [-4, 4] \cap \mathbb{Z}\). Some would stop here, and that would be wrong.

We need to come up with stronger proof. Our findings, \(k \in [-4, 4] \cap \mathbb{Z}\) were based on specific values for \(a,b,c,d\).

One famous inequality in mathematics is the inequality of arithmetic and geometric means:

\[(\frac{x_1^n+x_2^n+...+x_n^n}{n})^{\frac{1}{n}} \ge ... \ge \frac{x_1+x_2+...+x_n}{n} \ge (x_1 * x_2 * ... * x_n)^{\frac{1}{n}}\]If you are good at spotting patterns, you will see how it resembles our inequality.

\[\begin{cases} x_1 \rightarrow \lvert a \rvert \\ x_2 \rightarrow \lvert b \rvert \\ x_3 \rightarrow \lvert c \rvert \\ x_4 \rightarrow \lvert d \rvert \end{cases}\]So, if, we consider this:

\[(\frac{\lvert a \rvert ^4 + \lvert b \rvert ^4 +\lvert c \rvert ^4+\lvert d \rvert ^4}{4})^{\frac{1}{4}} \ge (\lvert a \rvert * \lvert b \rvert * \lvert c \rvert * \lvert d\rvert)^{\frac{1}{4}} \Leftrightarrow \\ a^4+b^4+c^4+d^4 \ge 4 * \lvert abcd \rvert\]- If \(abcd \ge 0\)
- Then \(a^4+b^4+c^4+d^4 + k*abcd \ge 4*abcd + k*abcd\);
- But \(abcd*(k+4) \ge 0\), for \(k \in [-4,4] \cap \mathbb{Z}\);

- If \(abcd \lt 0\)
- Then \(a^4+b^4+c^4+d^4 + k*abcd \ge -4*abcd + k*abcd\);
- But \(-abcd(4-k) \ge 0\), for \(k \in [-4, 4] \cap \mathbb{Z}\).

This tells us that the inequality works \(\forall k \in -4, 4 \cap \mathbb{Z}\).

The first observation you should make is that \(2021=43*41\).

When we have problems like this, it’s worth checking if using the two fundamental inequalities, AM-GM inequality and Cauchy–Bunyakovsky–Schwarz helps us.

Even if not fully obvious in our case, the one that helps is the CBS:

\[(\sum_{i=1}^{n} a_i^2) * (\sum_{i=1}^{n}b_i^2) \ge (\sum_{i=1}^{n}a_i*b_i)^2 \\ \text{ equality holds if } \frac{a_i}{b_i}=k \text{, for } k \in \mathbb{R}\]For 3 pair of numbers, \((a_1, b_1), (a_2, b_2), (a_3, b_3)\), CBS looks like this:

\[(a_1*b_1 + a_2*b_2 + a_3*b_3)^2 \le (a_1^2 + a_2^2 + a_3^2) * (b_1^2 + b_2^2 + b_3^2)\]Firstly, we know that: \(2021 \le (1*x+(-3)y+6z+(-1)*t)^2\)

So, thinking in terms of the CBS inequality, why don’t we consider the following:

\[\begin{cases} a_1 \rightarrow x \\ a_2 \rightarrow y \\ a_3 \rightarrow z \\ a_4 \rightarrow t \\ \end{cases}\]and

\[\begin{cases} b_1 \rightarrow 1 \\ b_2 \rightarrow -3 \\ b_3 \rightarrow 6 \\ b_4 \rightarrow -1 \\ \end{cases}\]In this regard, we can write things like this:

\[2021 \le (x-3y+6z+t)^2 \le \underbrace{(1^2+(-3)^2+6^2+1^2)}_{47}*\underbrace{(x^2+y^2+z^2+t^2)}_{\le 43} \le 2021\]This means our expression, \((x-3y+6z+t)^2\) is squeezed between \(2021\) and \(2021\), so the equality holds true, so:

\[\frac{x}{1}=\frac{y}{-3}=\frac{z}{6}=\frac{t}{-1}=k \text{ , } k \in \mathbb{R}\]Or:

\[\begin{cases} x=k \\ y=-3k \\ z=6k \\ t=-k \end{cases} \text{ , } k \in \mathbb{R}\]If we use this substitution:

\[x^2+y^2+z^2+t^2 \le 43 \Leftrightarrow \\ k^2 + 9k^2+36k^2+k^2 \le 43 \\ 47k^2 \le 43 \\ k \le \sqrt{\frac{43}{47}}\]And then again:

\[(x-3y+6z-t)^2 \ge 2021 \\ (k+9k+36k+k)^2 \ge 43*47 \\ 47^2*k^2 \ge 43*47 \\ k^2 \ge \frac{43}{47} \\ k \ge \sqrt{\frac{43}{47}}\]So we can safely assume that our \(k=\sqrt{\frac{43}{47}}\).

Now it’s easy to compute the expression: \(\lvert x^2 + y^2 + z^2 + t^2 \rvert=3 * \sqrt{\frac{43}{47}}\).

For 6 numbers: \(x_1 \le x_2 \le x_3\) and \(y_1 \le y_2 \le y_3\) the Rearrangement inequality can be written as:

\[x_1*y_2 + x_2*y_3+x_3*y_1 \le x_1*y_1 + x_2*y_2 + x_3*y_3\]If we pick:

\[\begin{cases} x_1=y_1=a \\ x_2=y_2=b \\ x_3=y_3=c \\ \end{cases}\]We obtain:

\[ab+bc+ca \le a^2 + b^2 + c^2 (*)\]Then:

\[2(ab+bc+ca) + ab+bc+ca \le 2(ab+bc+ca) + a^2 + b^2 + c^2 \Leftrightarrow \\ 3(ab+bc+ca) \le (a+b+c)^2 (**)\]Now, let’s get to our problem. We will compute the \(\Delta\) for \(x^2 + (a+b+c)x + k(ab+bc+ca) = 0\):

\[\Delta=(a+b+c)^2-4k(ab+bc+ca)\]For our equation to have solutions in \(\mathbb{R}\), we need to find \(k\) so that \(\Delta \ge 0\).

\[\Delta = \underbrace{(a+b+c)^2}_{\ge 3(ab+bc+ca)} - 4k(ab+bc+ca) \ge 0 \\\]Using (**), we can write:

\[\Delta \ge 3(ab+bc+ca) - 4k(ab+bc+ca) \ge 0 \Leftrightarrow \\ (3-4k) \ge 0 \Rightarrow k \le \frac{3}{4}\]Hermite’s Identity states that:

\[\sum_{k=0}^{n-1}[x+\frac{k}{n}] = [nx] \text{ , } \forall x \in \mathbb{R} \text{ ,and } n \in \mathbb{N}\]One idea is to make our existing terms (e.g. [\(\frac{x+3}{6}\)]) resemble the terms to Hermite’s identity (\([x+\frac{k}{n}]\)). In this regard we need to find a way to “isolate” the \(x\) outside the fraction(s).

One idea is to perform the following substitution:

\[y=\frac{x+1}{6}\]So our identity becomes:

\[[y+\frac{1}{3}] - [y+\frac{1}{2}] + [y+\frac{2}{3}] = [3y] - [2y] \text{ (*)}\]But:

\[\begin{cases} [3y] = [y] + [y+\frac{1}{3}] + [y+\frac{2}{3}] \\ [2y] = [y] + [y+\frac{1}{2}] \end{cases} \text{ (**)}\](*) and (**) proves the identity to be correct.

Even if it’s not obvious, let’s start again with the CBS inequality:

\[(\sum_{i=1}^{n} a_i^2) * (\sum_{i=1}^{n}b_i^2) \ge (\sum_{i=1}^{n}a_i*b_i)^2 \\\]If we pick \(b_1=b_2=b_3=...=b_n=1\) the inequality becomes:

\[(\sum_{i=1}^{n} a_i^2) * n \ge (\sum_{i=1}^{n}a_i)^2\]Expanding the sums:

\[(a_1+a_2+...+a_n)^2 \le n*(a_1^2 + a_2^2 + ...+ a_n^2)\]\(a_k \in \mathbb{R}_{+}\) so we can conclude:

\[a_1+a_2+...b_n \le n\]**12.** ^{easy} If \(a^2+b^2+c^2=3\) prove (\(\lvert a \rvert + \lvert b \rvert + \lvert c \rvert - abc) \le 4\), where \(a,b,c \in \mathbb{R}\). ^{Hint & Answer}

The following is true (\(\forall a,b,c \in \mathbb{R}\)):

\[a^2 + b^2 + c^2 = \lvert a \rvert^2 + \lvert b \rvert^2 + \lvert c \rvert^2 = 3\]Applying Cauchy–Bunyakovsky–Schwarz:

\[\underbrace{(a^2 + b^2 + c^2)}_{=3} * \underbrace{(1^2 + 1^2 + 1^2)}_{=3} \ge (\lvert a \rvert*1 + \lvert b \rvert*1 + \lvert c \rvert*1)^2\]From this \(\Rightarrow 3 \ge (\lvert a \rvert + \lvert b \rvert + \lvert c \rvert)\) (*).

Applying AM-GM inequality:

\[\frac{a^2+b^2+c^2}{3} \ge (abc)^{\frac{2}{3}}\]We can safely observe \(-abc \le 1\) (**).

(*) and (**) \(\Rightarrow\):

\[\underbrace{\lvert a \rvert + \lvert b \rvert + \lvert c \rvert}_{\le 3} \underbrace{- abc}_{\le 1} \le 4 \text{ is TRUE}\]Looking at:

\[\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2} \ge 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]We can group things in the following manner:

\[\underbrace{\frac{a}{b^2}+\frac{b}{a^2}}_{*} + \underbrace{\frac{b}{c^2}+\frac{c}{b^2}}_{**} + \underbrace{\frac{a}{c^2} + \frac{c}{a^2}}_{***} \ge \underbrace{\frac{1}{a} + \frac{1}{b}}_{*} + \underbrace{\frac{1}{b} + \frac{1}{c}}_{**} + \underbrace{\frac{1}{a} + \frac{1}{c}}_{***}\]There’s a pattern here!! If we manage to solve this inequality: \(\frac{m}{n^2} + \frac{n}{m^2} \ge \frac{1}{n} + \frac{1}{m}\), \(m,n \in \mathbb{R}_{+}^*\) we will be able solve our problem.

So let’s solve this:

\[\frac{m}{n^2} + \frac{n}{m^2} \ge \frac{1}{n} + \frac{1}{m} \Leftrightarrow \\\] \[m^3 + n^3 = mn(m+n) \ge 0 \Leftrightarrow \\\] \[m^2(m-n)-n^2(m-n) \ge 0 \Leftrightarrow \\\] \[(m-n)(m^2-n^2) \ge 0 \Leftrightarrow \\\] \[\underbrace{(m-n)^2}_{\ge 0} * \underbrace{(m+n)}_{\gt 0} \ge 0\]Now, we know the following:

\[\begin{cases} \frac{a}{b^2}+\frac{b}{a^2} \ge \frac{1}{a} + \frac{1}{b} \\ \frac{b}{c^2}+\frac{c}{b^2} \ge \frac{1}{b} + \frac{1}{c} \\ \frac{a}{c^2}+\frac{c}{a^2} \ge \frac{1}{a} + \frac{1}{c} \end{cases}\]If we sum all three inequalities, we’ve proven the original inequality.

If \(x^2+x \in \mathbb{Q}\) , then \(x^2+x=a \in \mathbb{Q}\)

If \(x^3+2x \in \mathbb{Q}\) , then \(x^3+2x=b \in \mathbb{Q}\).

Our purpose is to try expressing \(x\) using only \(a\) and \(b\) in a way we will prove \(x \in \mathbb{Q}\).

We start by doing some tricks with \(b=x^3+2x=x^3+\underbrace{(x^2-x^2)}_{=0}+\underbrace{(x-x)}_{=0}+2x\).

After regrouping terms, \(b\) becomes \(b=x\underbrace{(x^2+x)}_{=a}-\underbrace{(x^2+x)}_{=a} + x + 2x\).

So \(b=x(a+3)-a\), or \(a+b=x(a+3)\).

We are getting closer to the solution; the only thing remaining is to check if \(a=-3\).

Let’s suppose \(a=-3\), then \(x^2+x+3=0\), but this is impossible because the solutions \(x_1,x_2 \notin \mathbb{R}\). So, we can safely assume \(a \neq 3\).

If \(a \neq 3\), then we can write \(x=\frac{a+b}{a+3}\). But both \(a+b \in Q\) and \(a+3 \in Q\), then for sure \(x \in Q\).

The fact that \(3,5,7,11,13\) are all prime numbers is just a coincidence, but congratulations if you spot that.

Sometimes, it is easier to disprove, than to prove something, so ad absurdum let’s suppose \(a \ge b\).

Firstly, for the function \(f(x)=a^x\), if \(a\) is a constant greater than 1, we say that the function is increasing, meaning the value of \(f(x)\) increases with \(x\). Secondly, if \(a \lt 1\), then we say the function is decreasing, meaning the value of \(f(x)\) decreases while \(x\) increases.

That being said, if \(a \ge b\), then \(13^a \ge 13^b\) and \(5^a \ge 5^b\).

This means that \(3^a + 13^a \ge 17^a\), or \((\frac{3}{17})^a + (\frac{13}{17})^a \ge 1\).

Let \(g(x) : \mathbb{R} \rightarrow \mathbb{R}\), \(g(x)=(\frac{3}{17})^x + (\frac{13}{17})^x\). \(g(x)\) is stricly decreasing. Also, \(g(1)=\frac{16}{17} \lt 1\). But \(g(1) \lt g(a)\) thus, \(a \lt 1\) `(*)`

.

If our initial supposition is correct, then \(5^b+7^b \ge 11^b\). Following the same principle, we define \(h(x)=(\frac{5}{11})^b+(\frac{7}{11})^b \ge 1\) and we eventually conclude that \(b \gt 1\) `(**)`

.

`(*)`

and `(**)`

\(\Rightarrow\) our supposition is false, so \(a \lt b\) is true.

This problem is easier than it looks.

\[S=\underbrace{\frac{1}{u(2)*v(2)}}_{=\frac{1}{2*3}}+\underbrace{\frac{1}{u(3)*v(3)}}_{=\frac{1}{3*5}}+\underbrace{\frac{1}{u(4)*v(4)}}_{=\frac{1}{3*5}}+\underbrace{\frac{1}{u(5)*v(5)}}_{=\frac{1}{5*7}}+\underbrace{\frac{1}{u(6)*v(6)}}_{=\frac{1}{5*7}}+\underbrace{\frac{1}{u(7)*v(7)}}_{=\frac{1}{7*11}}...\]We see that the terms in our sum repeat themselves a number of times.

For example:

- \(\frac{1}{2*3}\) appears once;
- \(\frac{1}{3*5}\) appears two times;
- \(\frac{1}{5*7}\) appears two times;
- \(\frac{1}{7*11}\) will appear four times;
- … and so on.

If \(p, q\) are consecutive prime numbers, we define the set \(M=\{ n \in \mathbb{N} \lvert q \le n \lt p \}\). Th cardinal of \(M\) is exactly \(p-q\) for a given \(n\). It means that each term \(\frac{1}{qp}\) appears exactly \(p-q\) times in our sum.

With this in mind, we can re-write our sum as:

\[S=\frac{3-2}{2*3}+\frac{5-3}{5*3}+\frac{7-5}{5*7}+\frac{11-7}{7*11}+... \Leftrightarrow\] \[S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2011} \Leftrightarrow\] \[S=\frac{1}{2}-\frac{1}{2011}\]We use the AM-GM inequality for each of the following terms:

\[x^2+yz \le 2\sqrt{x^2yz} = 2x\sqrt{yz} \Leftrightarrow \\\] \[\frac{1}{x^2+yz} \le \frac{\sqrt{yz}}{2xyz} (*) \\\]But \(\sqrt{yz} \le \frac{y+z}{2} (**)\).

\((*), (**) \Rightarrow\):

\[\frac{1}{x^2+yz} \le \frac{\frac{y+z}{2}}{2xyz}\]Similarly, for the other terms:

\[\frac{1}{y^2+xz} \le \frac{\frac{x+z}{2}}{2xyz}\] \[\frac{1}{z^2+xy} \le \frac{\frac{x+y}{2}}{2xyz}\]If we sum everything up:

\[\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{\frac{y+z}{2}}{2xyz} + \frac{\frac{x+z}{2}}{2xyz} + \frac{\frac{x+y}{2}}{2xyz} \Leftrightarrow \\\] \[\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{1}{2}(\frac{x+y+z}{xyz}) \Leftrightarrow \\\] \[\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{1}{2} (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz})\]Let’s start by working on the expressions involving logarithms.

\[\begin{cases} a^x = bc \Leftrightarrow log_a(a^x) = log_a(bc) \Leftrightarrow x = log_a(b) + log_a(c) \\ b^y = ca \Leftrightarrow log_b(b^y) = log_b(ca) \Leftrightarrow y = log_b(c) + log_b(a) \\ c^z = ab \Leftrightarrow log_c(c^z) = log_c(ab) \Leftrightarrow c = log_c(a) + log_c(b) \\ \end{cases}\]Now, let’s change the base for our logarithms to a *common* number, \(m \in \mathbb{R}_{+}^{*}\):

Let’s define \(l_a = log_m(a)\), \(l_b = log_m(b)\), \(l_c = log_m(c)\).

We observe that, \(x,y,z\) can be written as:

\[\begin{cases} x=\frac{l_b+l_c}{l_a} \\ y=\frac{l_c+l_a}{l_b} \\ z=\frac{l_a+l_b}{l_c} \end{cases}\]The expression required to be proven becomes:

\[\frac{1}{2+\frac{l_b+l_c}{l_a}}+\frac{1}{2+\frac{l_c+l_a}{l_b}}+\frac{1}{2+\frac{l_a+l_b}{l_c}} \le \frac{3}{4} \Leftrightarrow\] \[\frac{l_a}{l_a + \underbrace{(l_a + l_b + l_c)}_{s_l}} + \frac{l_b}{l_b+\underbrace{(l_a+l_b+l_c)}_{s_l}} + \frac{l_c}{l_c+\underbrace{(l_a+l_b+l_c)}_{s_l}} \le \frac{3}{4} \Leftrightarrow\] \[1-\frac{l_a}{l_a+s_l}+1-\frac{l_b}{l_b+s_l}+1-\frac{l_c}{l_c+s_l} \ge 3-\frac{3}{4} \Leftrightarrow\] \[\frac{s_l}{l_a+s_l} + \frac{s_l}{l_b+s_l} + \frac{s_l}{l_c+s_l} \ge \frac{9}{4} \Leftrightarrow\] \[4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_b+s_l} + \frac{1}{l_c+s_l}) \ge 9\]Let’s suppose \(l_a \le l_b \le l_c\), also \(l_a>0\), then:

\[4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_b+s_l} + \frac{1}{l_c+s_l}) \ge 4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_a+s_l} + \frac{1}{l_a+s_l}) \ge 9\]So:

\[4*s_l(\frac{1}{l_a+s_l}) \ge 3 \Leftrightarrow\] \[1-s_l(\frac{1}{l_a+s_l}) \le \frac{1}{4} \Leftrightarrow\] \[\frac{l_a}{l_a+l_a+l_b+l_c} \le \frac{l_a}{4*l_a} \le \frac{1}{4}\]\(z=xy^2+x-1\), and \(z=2xy-1\).

We can write \(xy^2+x-1=2xy-1\). Eventually, this relationship becomes equivalent to \(x(y-1)^2\). But because \(x \neq 0 \Rightarrow y=1\).

If \(y=1\), then \(x=\frac{z+y}{2}\).

We pick a base \(d\) in the interval of \(a,b,c\). Basically we as for Problem 18.

After we do the substitution, we need to prove:

\[\frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z} \ge \frac{4x}{y+z} + \frac{4y}{z+x} + \frac{4z}{x+y}.\]Then, we need to prove that inequality holds true for:

\[\frac{y+z}{x} \ge \frac{4x}{y+z}\]Which is easy to prove using AM-GM inequality.

]]>Free suggestions in the beginning. If you follow all of them, you win. | |

Turn-Based Mode (the sinusoid doesn’t drop automatically) |

Free suggestions in the beginning. If you follow all of them, you win. | |

Turn-Based Mode (the sinusoid doesn’t drop automatically) |

^{(Source code)}

Controls

- To increase the angular frequency, \(\omega\), press:
`s`

; - To decrease the angular frequency, \(\omega\), press:
`x`

; - To increase the amplitude, \(A\), press:
`a`

; - To decrease the amplitude, \(A\), press:
`z`

; - To increase the phase: \(\varphi\), press:
`q`

; - To decrease the phase: \(\varphi\), press:
`w`

; - To
*drop*the sinusoid, press`p`

;

To win the game, you need to reduce the signal as close to zero as possible. It’s hard but not impossible. There’s a current threshold of `unit * 0.3`

. Surviving is not winning. The *Path of the Alternating Phases* is boredom.

You lose if the original signal spikes outside the game buffer (canvas).

A professional player turns off the suggestions, now enabled by default. If you are a savant, you can compute the *Fourier Series Coefficients* in your head. Cancel that noise!

The game was developed using p5js.

The source code (here) is not something I am particularly proud of.

Some discussion from around the web:

^{This game is a joke I put together during a weekend. I’m sorry for the graphics.}

The *newest* family member started walking and talking, which is/was delightful. My second *newest* family member started to pose simple questions requiring complicated answers, which is always challenging.

Software Engineering-wise, I am not precisely a *Professional* Software Engineer anymore since I switched to management years ago. But for reasons I am too afraid to admit, I still practice a technical hobby called *Recreational Programming*^{1}. This means that I program for fun in my limited spare time (a few hours a week); it makes me feel nostalgic for the years when programming was not a job.

In 2023, I didn’t achieve much and blogged little, but I did a few things I am proud of:

- I’ve finished and published an article I drafted and archived long ago called
*Demystifying bitwise_ operations in C*. Unexpectedly, It became pretty popular and got to the first page of hn, which is always lovely, but not a purpose in itself. I also had to pay 50$ to Netlify because of the traffic it created: I’ve forgot to compress the pictures and diagrams. Lesson learned. - I’ve curated and solved a list of leetcode (or leetcode-like) coding challenges. I haven’t yet published this list (it sits in an excel file), and I am not sure I will even publish it, because I don’t think it’s that useful for anyone other than me. Solving lots of exercises consistently through 2023 helped me improve my programming self-esteem. I’ve also recapped concepts I haven’t touched in years (e.g., tries, heaps, backtracking, and dynamic programming). I highly recommend you do the same, even if you don’t plan to interview at MAANG or similar. My list of exercises was built around this roadmap, so check that site if you don’t already know about it.
- I’ve finished reading this (free?!) book:
*THE NATURE OF CODE*by Daniel Shiffman. I had fun playing with graphics, vectors, forces and (the forces of) nature. In the*process*I haven’t learned “pure”*Processing*, but I am now comfortable writing p5js javascript code. I plan to publish some of my*modest*creations (2d and 3d) shortly. - I’ve watched (the pen & paper way) two online calculus courses as my knowledge of the subject began fading away. I also discovered two tools I wish were available when I was still in school: desmos and symbolab. They are so helpful. There’s no excuse for being terrible at math nowadays, except for lousy teachers, but you have YouTube to fix that.
- I found myself lost (again) in the rabbit hole called
*Fourier Analysis*. The last time I got caught there was in Uni, but out of desperation to pass exams, I lost myself in grinding math without understanding the subject at a deeper level. I still don’t**fully**get it, but at least I can*Fourier my way out of it*with a low pass filter. - I’ve switched from MAC to Linux at home and am amazed at how things have improved since 2012, when I last used Linux Mint. Everything works smoothly, and even if things are not as polished, I feel I’m in complete control of my
*computational*life. - I returned to using an RSS reader to get to valuable content hidden in the contemporary web’s noise.
- I have shown interest in AI and LLMs for work-related tasks (outside the area of creative programming). I also dabbled into cloud technologies, but that’s not something I can develop on my blog.

From a *cultural* perspective, 2023 was thin. I haven’t read much, or seen impactful movies (with a few notable exceptions):

- I was gifted the Seven Black Books of Carl Gustav Jung. Footnotes were essential for me to understand what the author had to say or what the editors thought they understood. Seeing how gifted people perceive and approach the miracle of existence is always interesting. In his quest for knowledge, Jung comes up with unique insights. I wish his journals were less heavy on unfamiliar (to myself) symbolism and less open to interpretations. I did find this YouTube Channel explaining quite a lot. All in all, I am conflicted: on the one hand, I’m not too fond of mysticism; on the other, I am attracted to it like a
*moth into flame*.

Music-wise, in 2023, I experimented with different genres I hadn’t touched before. I am more of a prog-rock guy, but 2023 was the year of alternative rock, punk, and the *aesthetics of the 80s*. My favorite tracks of 2023 (as reported by Spotify are):

- Bahaus - All we ever wanted
- Peter Schilling - Major Tom (Coming Home)
- The Clash - Train in Vain (Stand by Me)
- The Lemonheads - Mrs. Robinson
- The Fall - Big New Prinz HD
- The Cure - Lovesong
- Sad Lovers & Giants - Lope
- Carl Barat - Grimaldi
- Pulp - Common People
- The Feelies - Loveless Love
- The Veils - Jesus For The Jugular
- Peter Murphy - I’ll fall with your knife
- Arctic Monkeys - Four Out Of Five
- Gang Of Four - Damaged Goods
- The Specials - Ghost Town
- Brant Bjork - Radio Mecca
- Babyshambles - Beg, Steal or Borrow
- The Libertines - Up The Bracket
- Queens of the Stone Age - The Vampyre of Time and Memory
- Morrisey - Rebels without applause
- Editors - An End Has A Start
- Iggy Pop - Loves Missing
- Rome - To Die Among Strangers
- Killing Joke - Eighties
- DIIV - Doused
- Stray Cats - Stray Cat Strut
- Lord Huron - The World Ender
- Joy Division - Love Will Tear Us Apart
- The Pigeon Detectives - Romantic Type
- Madensuyu - Give
- Something You Can’t See (Audio) - Desert Sessions Vol. 12
- Talking Heads - Psycho Killer
- Arcade Fire - No Cars Go
- Them Crooked Vultures - No One Loves Me and Neither Do I
- The Strokes - Heart In A Cage

^{1}I’ve first heard about *Recreational Programming* by watching @tsoding, and it made sense.

- The system is flawed
- The system is simple
- My first important exam - Capacitate (2001)
- The second important exam -
*Bacalaureat*(2005) - The third exam -
*Admiterea la Politehnica*(2005) - Conclusions

The Romanian public education system, which, by the way, is free for everyone (Universities included), has its flaws. However, it is still one of the best in the world for making a handful of kids proficient in mathematics and sciences. It’s a system of contrasts. On the one hand, you have a bunch of *Elite Schools* and Universities that put a lot of emphasis on scientific education, and on the other hand, you have the “average” Public Schools creating kids that can barely read, do simple arithmetics, or understand a simple text. Every middle-class parent that cares about the education of its children wants to send their children to an *Elite Public School*. The average school is actually below average, and Private Schools don’t have the best reputation, are too expensive, or are not that popular. The term *Elite* is not formalised, but there are “unofficial” rankings about how good the school is.

Getting there is mainly a meritocratic process. The child (or adolescent) must pass a few exams, and if he/she is 10/10 (or 9.50/10) he/she will be accepted almost anywhere. If you are 8.5/10, you will still get to a decent school. Lower than that, not so good. This state of affairs is similar to all the countries from the Eastern Bloc, not only Romania. So, if you speak with a Bulgarian, Hungarian, Polish, Ukrainian, or Russian person, they will tell you the same story.

This is the secret recipe for why Eastern and Central Europe is so well-represented at the International Math Olympiad. Those *Elite Schools*, spread in every region of the country, try to create the next generation of *olimpici* (kids that go to the Olympiads). Those *Elite Schools* are almost always *Math oriented*. It’s practically a *cult* (without any negative connotation).

Let’s look at the following table of results for IMO:

If you take Russia and Ukraine aside, Romania, Bulgaria, Hungary, and Poland have less than 80 million citizens together, but have accounted for 261 Gold Medals through the years, more than China or USA, the World’s two Superpowers. Of course, most of those students (and the ones that don’t qualify to the International but are way above the world’s average) leave Romania (or Poland, Hungary, and Bulgaria) to more prosperous, more developed countries. So what’s with all the fuss? *Brain drain* is a sad reality few try to counter.

The way those schools create the next-gen *olimpic* is simple: you make the kids take 4-6 of Math a week (or more), you give extensive Math homework (tens or hundreds of exercises per week), you ask the student to come to special classes designed to prepare them for the Olympiad, and then you repeat the formula for 12 years. Not everyone can make it, and the competition is tight. There’s a flaw here: some kids are differently wired than others and like challenges, but the constant grind is not suitable for everyone. Our Educational system is flawed because it expects almost everyone to ace Math. For example, I had colleagues in my class who are now lawyers, doctors, or work in creative industries but finished the *Elite (Math) School* only because the education there was so much better. So yes, in Romania, there are lawyers who studied Riemann Sums, and dentists who did prove Euler’s identity with Taylor Series. We did learn both concepts in high school, but those are not necessarily part of the standard curricula.

In the Romanian region where I was born (Oltenia), the elite high schools were: Colegiul National Carol I, Colegiul National Fratii Buzesti, and Colegiul National Elena-Cuza (but historically speaking, this one was focused on Humanities). I’ve picked CNC1, but the alternative would’ve been as good. I remember that one of the role models we had in our first year of high school was Mihai Patrascu (he was 3-4 years older than us), but all the teachers told us how great he is/was as a student. He was great at Computer Science and had fantastic results in Math.

In Bucharest, you can probably find even better schools, as the competition between students is more fierce. Everyone knows about Colegiul National Gheorghe Lazar, Colegiul National Sfantul Sava, or Colegiul National Tudor Vianu, a school that only in the last decade had 30 gold medalists, 53 silver medalists and 36 bronze medalists in various international competitions (mainly Math, Physics and Informatics).

The (sub)cult(ure) of Mathematics in Eastern Europe is older than communism, and the results were always notable. Communists loved Mathematics a lot. It was one of the few occupations that could instantly reward you with a decent life. It didn’t require you *to scream* total allegiance to the Communist Party, and it allowed you to have a few excentricities well tolerated by a dictatorial regime. So people went for it. To put it like this, Nicolae Ceasusescu’s daughter, Zoia Ceausescu, was an accomplished mathematician; even if her parents disagreed with her passion, eventually, they accepted this endeavor to be worthy of a *Communist Princess*. The regime did not inflate her skills in mathematics; for example, a well-regarded academician Solomon Marcus, vouched for her talent in an interview long after communism fell and Zoia died. He had no incentive to lie about the subject; it was a free country.

As a fun fact, the current mayor of Bucharest, Nicusor Dan is a two-time Gold Medalist at the International Mathematics Olympiad. He had a perfect score each time.

The Romanian education system is simple and shows little flexibility. It doesn’t care about too much about the children’s talents

You cannot take “extra classes”, and the first eight years of school are standard for everyone regardless of their skills and aptitudes:

- The first cycle is called
*Scoala Primara*(Primary school).*Scoala Primara*is about four years (or five) years long. You start school when you are 6, 7, or 8. I started school at 6 (in 1993), but most people send their kids to school at around 7. - The second cycle,
*Scoala Generala*, takes four years and starts when you are around 11-12. In recent times, if you want to attend an*Elite School*, you must pass an admission exam. The exam is not standard, and the*Elite Schools*decide the curricula. Sometimes, the*curricula*contain topics not part of the*formal curricula*, so the aspiring kid has to take private tutoring to prepare for this exam. - At the end of the
*Scoala Generala*, you must pass another exam called*Capacitate*(in my time) or*Teste Nationale*(in recent times). This exam is the first important one. If you are close to 10/10, you will be admitted to the*Elite School*; anything under 9/10 makes it more difficult or impossible. The admission process is totally transparent. You cannot bribe your way into admission, or at least not in this phase of the process. - The third cycle is
*Liceul*. This is where diversification first appears. You can go to a*Profil Real*to study*matematica-informatica*or a*Profil Uman*, to study*Humanities*. If you pick*Profil Uman*, it means your mathematical education stops. Usually, the best pick is*matematica-informatica*, which can also be spiced up with additional hours of math, physics, or informatics (basically Computer Science). - At the end of high school, there’s a second exam called Baccaluareat (like in France). Getting a big mark at Baccalaureat helps you get admitted to local Universities, but there’s also an admission exam for the really good ones.

It was 2001, and the first important Math exam I had to take was called *Capacitate*. I was 13 or 14 years old. Thanks to the internet, I did find the subjects online, so I had to solve them almost 23 years apart.

The exam was tailored to make it easy to get a mark higher than 7/10 and significantly more challenging to get a mark higher than 9/10.

Anything better than 9/10 would help you land a place in an *Elite* school. This was the target, so I aimed to get a perfect score (I was close: 9.80/10, if I remember correctly). Since I was ten years old, I have also participated in Math Olympiads, always making it into the top 25 students of my region (my best was 4th rank), so I remember *Capacitate* to be manageable.

In *Scoala Generala* I remember a lot of emphasis was put on basic Algebra and Geometry (2d and 3d). The student had to develop the *graphical* intuition of things.

The subjects for the *Capacitate* were the following. We had two hours to solve them:

- The result of \(5*6-7*4\) is:
`...`

- The obvious answer is \(30-28=2\).

- Between \(6548\) and \(2145\), the number divisible by 3 is
`...`

- The rule is simple: for a number to be divisible by 3 the sum of its digits is divisible by 3, so the answer is: \(2145\). (\(2+1+4+5=12\), \(12\) is divisible by \(3\)).
- To solve this exercise, there’s no need to perform the division.

- The arithmetic mean of the numbers \(18\) and \(12\) is
`...`

- This exercise simply asses if the kid understands the concept of an arithmetic mean, the answer is \(15\).

- Given the
*proportion*\(\frac{x}{6}=\frac{3}{2}\), \(x\) is`...`

- \(x\) is \(\frac{6*3}{2}=9\).

- Between \(a=5\sqrt{2}\) and \(b=2\sqrt{13}\) which one is bigger ?
- The answer is \(2\sqrt{13}\). Back in school, I had to remember a few radicals to use them as needed. \(\sqrt{13}\) was one of them. I have forgotten them by now. We also had to learn the algorithm to compute any \(\sqrt{}\) as needed.

- An isosceles triangle \(\text{ABC}\) with \([AC]\equiv[AC]\) has the angle \(\measuredangle \text{ABC}=35^{\circ}\), what is \(\measuredangle \text{BAC}=\)
`...`

?- The sum of the angles in a triangle is \(180^{\circ}\), and we know that in an isosceles triangle the two angles \(\measuredangle \text{ABC}\) and \(\measuredangle \text{ACB}\) are equal. So the answer is \(\measuredangle \text{BAC}=180^{\circ}-2*35^{\circ}=110^{\circ}\).

- A rhombus has the diagonals \(10\) and \(24\)?
- What is the length of one of its sides?
- The side is \(\sqrt{(\frac{10}{2})^{2}+(\frac{24}{2})^{2}}=13\)

- What is the perimeter of the rhombus?
- Now that we know one side, we multiply its value by \(4*13=52\).

- What is the length of one of its sides?
- A square prism has it’s volume \(80cm^3\), and the height \(5cm^2\)
- The base surface of the prism is
`...`

- Applying the formula, we get the answer \(16cm^2\)

- The side of the prim’s base is
`...`

- Applying the formula, we get \(4cm^2\)

- The base surface of the prism is
- A cube has one of its sides \(2cm\). The total area of the cube is
`...`

- The answer is \(24cm^2\)

- A specific product had a price increase of 10%. After a while, it got a new 10% price increase, and it now costs \(133100 \text{lei}\).
- What was the initial price of the product?
- \(x\) is the initial price of the product;
- \(a=x+\frac{10x}{100}\) is the price of the product after then first 10 percent increase;
- After the second price increase \(a+\frac{10a}{100}=133100\);
- We compute \(a=133100*\frac{100}{110}=121000\);
- We compute \(x=121000*\frac{100}{110}=110000\).

- What is the total price increase in percentages?
- We need to solve this equation \(110000+110000\frac{x}{100}=133100\);
- We determine the total price increase to be \(\frac{21}{100}=21\)%;

- What was the initial price of the product?
- Considering the following functions \(f : \mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=ax+b-9\) and \(g : \mathbb{R} \rightarrow \mathbb{R}\), \(g(x)=2bx-a\) and \(a,b \in \mathbb{R}\):
- Determine \(a,b\) knowing that \(A(2,3)\) belongs to \(f(x)\) and \(g(x)\);
- If \(A(2,3)\) belongs to the \(f(x)\) and \(g(x)\), then the following is true: \(f(2)=a*2+b-9=3\) and \(g(2)=4b-a=3\).
- After solving the system of equations \(a=5\) and \(b=2\).

- For \(a=5\) and \(b=2\) plot \(f(x)\) and \(g(x)\).
- Substituting the values, we get the functions: \(f(x)=5*x+2-9=5*x-7\) and \(g(x)=4*x-5\)
- To plot them, we need to understand where the functions intersects \(OX\) and \(OY\).
- \(f(0)=-7\) so the \(OY\) axis is intersected by \(f(x)\) at \((0, -7)\);
- \(5x-7=0\), we get \(x=\frac{7}{5}\), so the \(OX\) axis is intersected at \((\frac{7}{5}, 0)\);
- \(g(0)=-5\), so the \(OY\) axis is intersected by \(g(x)\) at \((0, -5)\);
- \(4x-5=0\), we get \(x=\frac{5}{4}\), so the \(OY\) axis is intersected at \((\frac{5}{4}, 0)\);
- We plot

- For \(a=5\) and \(b=2\), \(f\) intersects \(OY\) in \(B\), and \(g\) intersects \(OY\) in \(C\). Compute the distances from \(C\) to \(AB\).
- We know that \(A\) is on \(f(x)\) and so is \(B\). So we can safely assume that the line equation for \(AB\) is the given by \(f(x)\), so \(y-5x+7=0\).
- A student can also compute the line equation from two points \(A\) and \(B\) with the formula: \(\frac{x-x_A}{x_B-x_A}=\frac{y-y_A}{y_B-y_A}\), but it’s a little bit overkill.
- We have to compute the distance from \(C(0,-5)\) to \(y-5x+7=0\) or \(-5x+y+7=0\).
- There’s a formula for that, \(d(ax+by+c, P(x_p, y_p))=\frac{\vert ax_p+by_p+c \vert}{\sqrt{a^2+b^2}}\).
- Substituting the values in the formula \(d=\frac{\vert (-5)*(0) + 1 * (-5) + 7 \vert}{\sqrt{(-5)^2 + 1^2}}=\frac{2}{25+1}=\frac{\sqrt{26}}{13}\)

- Determine \(a,b\) knowing that \(A(2,3)\) belongs to \(f(x)\) and \(g(x)\);

- Plot the Frustum of a Right Circular Cone. The volume is \(312\pi \text{cm}^3\), and it’s height is \(8\text{cm}\). We section the frustum with a plane parallel to its base; this plane is going straight through the middle of its height. The surface of this section is \(36\pi \text{cm}^2\).
- The radius of the smaller base of the frustum is \(r\). Prove that \(r^2 - 12*r + 27=0\).
- The vormula for the Volume is \(V=\frac{1}{3}h\pi (R^2 + r^2 + Rr)\).
- If we substitute the values, we obtain \(R^2+r^2+Rr=117\).
`(1)`

- If the surface of the section is \(S_{s}=36\pi=\pi*{r_s}^2\), the radius of the section is \(r_s=\sqrt{36}=6\).
- But, \(\frac{R+r}{2}=r_s\), so we can say that \(R+r=12\).
`(2)`

- Using
`(1)`

in`(2)`

, we obtain \(R^2+r(R+r)=(12-r)^2 + r*12=117\). After a few more steps we prove that \(r^2-12r+27=0\).

- Determine the radiuses of the two bases of the frustum.
- We already know the relationship: \(r^2-12r+27=0\) holds true, so we can compute the radiuses as \(3\) and \(9\).

- If the radiuses are \(9\) and \(3\), compute the sector of the arc, which is the lateral unfolding of the frustum.
- There’s a formula for this \(A_l = \pi G(R+r)\), and \(G=\sqrt{h^2+(R-r)^2}=10\). So \(A_l=120\pi \text{cm}^2\).

Just like *Capacitate*, *Bacalaureat* is tailored in a way you can pass it (the minimum requirement is 5/10), but it makes it harder to score 10/10. Again, I couldn’t score 10/10, but 9.90/10 was close.

In Romania, if you opt for a *matematica-informatica* specialization, you study pieces of everything: some advanced algebra, basic statistics, linear algebra, descriptive geometry, and lots of real analysis.

The high school math syllabus is quite intense, or at least this is how I’ve felt it to be. I had to allocate lots of extra time to grasp the concepts my brilliant teacher constantly threw at me (thanks, Mrs. Georgescu, you were great!).

On the one hand, I am grateful I did a lot of calculus in high school; it helped me with math and physics courses during my University years; on the other hand, I feel that making calculus mandatory for the Baccalaureat punishes too harshly the students who are not mathematically inclined.

The following paragraphs contain a rough translation of the High-School Math Syllabyus (the version from 2017) for the *matematica-informatica* specialization:

**Sets and Elements of Mathematical Logic**(First Year)- Set of real numbers: algebraic operations with real numbers, ordering of real numbers, absolute value of a real number, approximations by deficiency or by excess, integer part, fractional part of a real number; operations with intervals of real numbers.
- Proposition, predicate, quantifiers
- Basic logical operations (negation, conjunction, disjunction, implication, equivalence), correlated with operations and relations between sets (complement, intersection, union, inclusion, equality); reasoning by
*reductio ad absurdum*. - Mathematical induction

**Sequences**(First Year)- Ways to define a sequence, bounded sequences, monotonic sequences
- Specific sequences: arithmetic progressions, geometric progressions, the formula for the general term in terms of a given term and ratio, the sum of the first n terms of a progression
- Conditions for n numbers to form an arithmetic or geometric progression

**Functions; Graphical Readings/Representation**(First Year)- Cartesian reference system, Cartesian product; representation through points of a Cartesian product of numerical sets; algebraic conditions for points located in quadrants; lines in the plane of the form \(x=m\), or \(y=m\) with \(m \in \mathbb{R}\).
- Function: definition, examples, examples of correspondences that are not functions, ways to describe a function, graphical readings. Equality of two functions, the image of a set through a function, the graph of a function, restrictions of a function;
- Numerical functions \((F={f:D \rightarrow \mathbb{R}, D \subseteq \mathbb{R}})\); geometric representation of the graph: intersection with the coordinate axes, graphical solutions of equations and inequalities of the form \(f(x)=g(x), (\gt, \lt, \geq, \leq)\); properties of numerical functions introduced through graphical reading: boundedness, monotonicity; other properties: parity/oddity, symmetry of the graph with respect to lines of the form \(x=m\), \(m \in \mathbb{R}\), periodicity;
- Function composition

**First-Degree Function**(First Year)- Definition; graphical representation of the function: \(f:\mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=ax+b\), where \(a, b \in \mathbb{R}\)
- Intersection of the graph with the coordinate axes
- Graphical interpretation of the algebraic properties of the function: monotonicity and the sign of the function; studying monotonicity through the sign of the difference \(f(x_{1})-f(x_{2})\) (or by studying the sign of the ratio: \(\frac{f(x_1)-f(x_2)}{x_{1}-x_{2}})\), \(x_1, x_2 \in \mathbb{R}, x_1 \neq x_2\).
- Inequalities of the form \(ax+b (\gt, \lt, \geq, \leq) 0\) studied on or over intervals of real numbers.
- Systems of first-degree inequalities

**Second-Degree Function**(First Year)- Graphical representation of the function: \(f:\mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=ax^2+bx+c\), \(a,b,c \in \mathbb{R}, a \neq 0\), intersection of the graph with the coordinate axes, equation \(f(x)=0\), symmetry with respect to lines of the form \(x=m\), \(m \in \mathbb{R}\)
- Viète’s relations

**Geometric Interpretation of the Algebraic Properties of the Second-Degree Function**(First Year)- Monotony; studying monotony through the sign of the difference \(f(x_1)-f(x_2)\) or through the rate of increase/decrease: \(\frac{f(x_1)-f(x_2)}{x_1-x_2}\), extreme point, vertex of the parabola;
- Positioning of the parabola relative to the x-axis, the sign of the function, inequalities of the form \(ax^2+bx+c(\gt, \lt, \geq, \leq) 0\), \(a, b, c \in \mathbb{R}, a \neq 0\), on or over intervals of real numbers, geometric interpretation: images of intervals (projections of portions of a parabola onto the y-axis);
- Relative position of a line with respect to a parabola

**Vectors in the Plane**(First Year)- Oriented segment, vectors, collinear vectors
- Vector operations: addition (triangle rule, parallelogram rule), properties of the addition operation; scalar multiplication, properties of scalar multiplication; collinearity condition, decomposition along two non-collinear vectors

**Collinearity, concurrency, parallelism - vector calculus in plane geometry**(First Year)- Position vector of a point
- Position vector of the point that divides a segment in a given ratio, Thales’ theorem (conditions for parallelism)
- Position vector of the centroid of a triangle (concurrency of medians of a triangle)
- Menelaus’ theorem, Ceva’s theorem

**Elements of Trigonometry**(First Year)- Trigonometric circle, definition of trigonometric functions;
- Reduction to the first quadrant; trigonometric formulas

**Applications of Trigonometry and the Scalar Product of Two Vectors in Plane Geometry**(First Year)- Scalar product of two vectors: definition, properties. Applications: cosine theorem, conditions of perpendicularity, solution of right-angled triangles
- Vector and trigonometric applications in geometry: sine theorem, solution of arbitrary triangles
- Calculation of the radius of the inscribed and circumscribed circles in a triangle, calculation of lengths of important segments in a triangle, calculation of areas

**Sets of Numbers**(Second Year)- Real numbers: properties of powers with rational, irrational, and real exponents of a positive non-zero number, rational approximations for real numbers
- \(n\)th root (\(n \in \mathbb{R}\)) of a number, properties of radicals;
- Notion of logarithm, properties of logarithms, calculations with logarithms, logarithmic operation
- Set of complex numbers in algebraic form, conjugate of a complex number, operations with complex numbers. Geometric interpretation of addition and subtraction operations of complex numbers and their multiplication by a real number
- Solving a quadratic equation with real coefficients. Quadratic equations

**Functions and Equations**(Second Year)- Power function with a natural exponent: \(f:\mathbb{R} \rightarrow D\), \(f(x)=x^n\), \(n \in \mathbb{N}\), \(n \geq 2\)
- Radical function: \(f:\mathbb{R} \rightarrow D\), \(f(x)=(x)^{\frac{1}{n}}\), \(n \in \mathbb{N}\), \(n \geq 2\), \(D=[0, \infty)\), \(n\) even or odd;
- Exponential function: \(f:[0, \infty)\), \(f(x)=a^x\), \(a \ge 0\), \(a \neq 1\);
- Logarithmic function: \(f:[0, \infty]\), \(f(x)=log_{a}(x)\), \(a \ge 0\), \(a \neq 1\);
- Injectivity, surjectivity, bijectivity; invertible functions: definition, graphical properties, necessary and sufficient condition for a function to be invertible
- Direct and inverse trigonometric functions

**Counting Methods**(Second Year)- Ordered finite sets. Number of functions \(f:A \rightarrow B\), where \(A, B\) are finite Sets;
- Permutations:
- Number of ordered sets obtained by arranging a finite set with \(n\) elements
- Number of bijective functions \(f:A \rightarrow B\), where \(A,B\) are finite Sets;

- Arrangements
- Number of ordered subsets with k elements, \(k \leq n\), that can be formed with \(n\) elements of a finite Sets;
- Number of injective functions \(f:A \rightarrow B\), where \(A, B\) are finite Sets;

- Combinations
- Number of subsets with k elements, \(0 \leq k \leq n\), of a finite Sets with \(n\) elements. Properties: complementary combinations formula, number of all subsets of a set with \(n\) elements

- Newton’s Binomial

**Financial Mathematics**(Second Year)- Elements of financial calculations: percentages, interest, VAT
- Collection, classification, and processing of statistical data: statistical data, graphical representation of statistical data
- Interpretation of statistical data through position parameters: means, variance, deviations from the mean
- Equally likely random events, operations with events, probability of a compound event composed of equally likely events

**Geometry**(Second Year)- Cartesian coordinate system in the plane, coordinates of a vector in the plane, coordinates of the sum of vectors, coordinates of the product between a vector and a real number, Cartesian coordinates of a point in the plane, distance between two points in the plane
- Equations of a line in the plane determined by a point and a given direction, and equations of a line determined by two distinct points
- Conditions for parallelism, conditions for the perpendicularity of two lines in the plane; calculation of distances and areas

**Matrices and Systems of Linear Equations**(Third Year)- Matrices;
- Matrix operations: addition, multiplication, multiplication of a matrix by a scalar, properties;
- Determinants;
- Systems of Linear Equations;
- Invertible matrices for \(n \leq 4\);
- Matrix equations;
- Linear systems with at most 4 unknowns, Cramer’s systems, matrix rank
- Study of compatibility and solution of systems: Kronecker-Capelli property, Rouchè property, Gaussian method
- Applications: equation of a line determined by two distinct points, area of a triangle, and collinearity of three points in the plane

**Elements of Real Analysis**(Third Year)- Elementary notions about sets of points on the real line: intervals, boundedness, neighborhoods, closed line, symbols ∞ and −∞
- Real functions of a real variable: polynomial function, rational function, power function, radical function, logarithmic function, exponential function, direct and inverse trigonometric functions
- Limit of a sequence using neighborhoods, convergent sequences
- Monotony, boundedness, limits; the squeeze theorem
- Weistrass Property
- The number \(e\)
- Limits in the form \(((1+u_n)^\frac{1}{u_n})\), \(u_n \rightarrow 0\), \(u_n \neq 0\), \(n \in N\);
- Operations with Sequences that have a Limit
- Limits of functions: graphical interpretation of the limit of a function at a point using neighborhoods, one-sided limits
- Calculation of limits for the studied functions; exceptional cases in the calculation of limits of functions
- Asymptotes of the graph of studied functions: vertical asymptotes, oblique asymptotes

**Function Continuity**(Third Year)- Continuity of a function at a point in its domain, continuous functions, graphical interpretation of the continuity of a function, studying continuity at points on the real line for the studied functions, operations with continuous functions
- Darboux’s property, the sign of a function continuous on an interval of real numbers, studying the existence of solutions to equations in \(\mathbb{R}\);

**Differentiability**(Third Year)- Tangent to a curve, derivative of a function at a point, differentiable functions, operations with differentiable functions, calculation of first and second-order derivatives for the studied functions
- Functions differentiable on an interval: extreme points of a function, Fermat’s theorem, Rolle’s theorem, Lagrange’s theorem and their geometric interpretation, the corollary of Lagrange’s theorem regarding the derivative of a function at a point
- The role of the first derivative in the study of functions: monotony of functions, extreme points
- The role of the second derivative in the study of functions: concavity, convexity, inflection points
- L’Hôpital’s rules

**Graphical Representation of Functions**(Third Year)- Graphical representation of functions
- Graphical solution of equations, using the graphical representation of functions to determine the number of solutions to an equation
- Graphical representation of conics (circle, ellipse, hyperbola, parabola)

**Advanced Algebra**(Third Year)- Groups
- Internal composition law (algebraic operation), operation table, stable part
- Group, examples: numerical groups, matrix groups, permutation groups, additive group of residue classes modulo \(n\)
- Subgroup
- Finite group, operation table, order of an element
- Morphism, group isomorphism

- Rings and Fields
- Ring, examples: numerical rings (\(Z, Q, R, C\)), \(Z_{n}\), matrix rings, rings of real functions
- Field, examples: numerical fields (\(Q, R, C\)), p-primary
- Ring and field morphisms

- Polynomial Rings with Coefficients in a Commutative Field (Q, R, C, \(Z_p\), p-primary)
- Algebraic form of a polynomial, polynomial function, operations (addition, multiplication, scalar multiplication)
- Remainder theorem; polynomial division, division by \(X−a\), Horner’s scheme
- Polynomial divisibility, Bézout’s theorem; greatest common divisor and least common multiple of polynomials, factorization of polynomials into irreducible factors
- Roots of polynomials, Viète’s relations
- Solving algebraic equations with coefficients in (\(Z, Q, R, C\)), binomial equations, quadratic equations, reciprocal equations

- Groups
**Real Analysis**(Fourth Year)- Antiderivatives of a function defined on an interval. Indefinite integral of a function, properties of the indefinite integral, linearity. Standard antiderivatives
- Definite Integral
- Subdivisions of an interval \([a,b]\), norm of a subdivision, system of intermediate points, Riemann sums, geometric interpretation. Definition of the integrability of a function on an interval \([a,b]\);
- Properties of the definite integral: linearity, monotony, additivity with respect to the integration interval.
- Leibniz-Newton formula
- Integrability of continuous functions, mean value theorem, geometric interpretation, theorem of the existence of primitives for a continuous function
- Methods of calculating definite integrals: Integration by parts, integration by change of variable.
- Calculation of integrals of the form \(\int_{a}^{b} \frac{P(x)}{Q(x)} dx\), using the method of partial fraction decomposition.

The *Bacalaureat* exercises I had to solve (in 3 hours) were the following:

- If \(f:\mathbb{R} \rightarrow \mathbb{R}\) is \(f(x)=x-3\), what is the value of the prouct \(f(1)*f(2)*...*f(7)\)?
- f(3)=0, so the product is itself 0. No need for additional computations.

- How many non-empty subsets of the set \(\mathbb{Z}_{3}\) have a sum of elements equal to 0̂?
- \(\mathbb{Z}_{3}\) refers to the set of integers modulo 3.
- We will have to consider various combinations and check the condition.
- The subsets are \(\{0\}\), \(\{1,2\}\), and \(\mathbb{Z}_{3}\), so the final answer is 3.

- If the function \(f:\mathbb{R} \rightarrow \mathbb{R}\) is \(f(x)=-x^4+2x\) what is the value of \((f \circ f)(1)\) ?
- We start with the definition \((f \circ f)(x)=f(f(x))=-(-x^4+2x)^4 + 2(-x^4+2x)\)
- We substitue \(x=1\), \((f \circ f)(1)=-1+2=1\)

- What is the probability that an element n from the set \(\{0, 1, 2, 3, 4\}\) satisfies the relationship \(2^n + 5^n = 3^n + 4^n\) ?
- The relantionship is
*satisfied*if \(n=0\) or \(n=1\), using the formula for probabilities, the answer is \(P=\frac{2}{5}\).

- The relantionship is
- How many real solutions does the \(x^4=16\) equation have?
- \((x^4)=(x^2)^2=4\), because \(x \in \mathbb{R}\), the real solutions are \(2\) and \(-2\), so the answer is 2.

Considering the function: \(f:\mathbb{R}\), \(f(x)=e^x+x+\frac{1}{2}\).

- Compute \(f'(x)\).
- We simply compute: \(f'(x)=(e^x+x+\frac{1}{2})' = e^x + 1\)

- Compute \(\int_{0}^{1}f(x)dx\).
- We simply compute: \(\int_{0}^{1}f(x)dx=\int_{0}^{1}(e^x+x+1)dx=(e^x+\frac{x^2}{2}+\frac{x}{2}) \Big\|_0^1=e\).

- How is the function \(f\) over the set of real numbers: convex or concave?
- We differentiate the function twice and check the conditions:
- If \(f''(x) \geq 0\), then the function is convex
- If \(f''(x) \leq 0\), then the function is concave

- We know \(f'(x)=e^x+1\), so we can easily compute \(f''(x)=e^x\).
- It’s a known fact that \(e^x \geq 0\) so the answer is: \(f\) is convex on \(\mathbb{R}\).

- We differentiate the function twice and check the conditions:
- What is the limit: \(\lim_{x \to 1} \frac{f(x)-f(1)}{x-1}\) ?
- Cheesy one. The definition of the derivative in \(a\) is: \(\lim_{x \to a} \frac{f(x)-f(a)}{x-a}\);
- We have to compute \(f'(1)=e^1+1=e+1\). The answer is \(e+1\).

- What is the limit \(\lim_{x \to \infty} \frac{\sqrt{n}}{n}\) ?
- We can do this simple trick: \(\lim_{n \to \infty} \frac{\sqrt{n}}{n}=\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n}*\sqrt{n}}=\frac{1}{\infty}=0\).

- What is the distance between the two points: \(A(1,3,5)\) and \(B(3,5,7)\)?
- We compute the distance: \(d=\sqrt{(3-1)^2+(5-3)^2+(7-5)^2}=\sqrt{3*2^2}=2*\sqrt{3}\)

- What is the radius of the circle: \(x^2+y^2=4\) ?
- The answer is \(r=2\) (fundamental equation of the circle)

- What is \(cos^2\pi + sin^2\pi\) ?
- The answer is 1, as per the fundamental: \(sin^2x+cos^2x=1\).

- What is the modulus of the complex number \(z=\frac{5+8*i}{8-5*i}\)?
- We need to compute: \(\vert z \vert\).
- If \(z=a+b*i\), then \(\vert z \vert=\sqrt{a^2+b^2}\)
- We need to reduce our \(z\) to a form were it’s easy to identity \(a\) and \(b\)
- In this regard, we can write \(z=\frac{(5+8*i)(8+5*i)}{(8-5*i)(8+5*i)}\)
- After doing all the computations: \(z=i\), so \(\vert z \vert=1\)

- What is the area of a triangle with its sides: \(3\), \(3\) and \(4\) ?
- The elegant way of doing it is to apply Heron’s theorem: \(A=\sqrt{p(p-a)(p-b)(p-c)}\), where \(p=\frac{a+b+c}{2}\).
- The answer is \(A=\sqrt{5*2*2*1}=2*\sqrt{5}\).

- What is the equation of the tangent line to the parabola \(y^2 = 2x\) passing through the point \(P(2, 2)\)?
- Applying the equation, we get to the form: \(2*y=x+2\).

Considering the matrices: \(I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), \(O_{2}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\), \(J=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), \(K=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), a matrix \(M \in M_{2}(\mathbb{R})\) is *nilpotent*, if there is \(n \in \mathbb{N}^{*}\), so that \(M^n=O_2\).

- Prove that \(O_2\) and \(J\) are nilpotent
- \(O_2^n=O_2\), \(∀ n \in \mathbb{N}^{*}\), so we can pick an arbitrary \(n\), and the nilpotency condition is satisfied.
- For \(J\) we just compute \(J^2\), and it becomes evident that the nilpotency condition is satisfied for \(n=2\).

- Show that the \(K\) matrix is neither inversible nor nilpotent.
- For a matrix to be invertible, \(det(A) \neq 0\), but \(det(K)=0\), so \(K\) is not invertible;
- We can prove \(K^n=K\), \(∀ n \in \mathbb{N}^{*}\), but \(K \neq O_2\), so \(K\) is not nilpotent;

- Prove that \(X \in M_2(\mathbb{R})\), \(X=\begin{pmatrix} p & q \\ r & s \end{pmatrix}\), satisfies the identity: \(X^2 - (p+s)X + (ps-rq)I_2=O_2\).
- First step is to compute \(X^2\).
- In this regard: \(X^2=\begin{pmatrix} p^2 + qr & pq + qs \\ rp + rs & s^2 + qr\end{pmatrix}\);
- The second step is to compute \((p+s)X=\begin{pmatrix} (p+s)p & (p+s)q \\ (p+s)r & (p+s)s \end{pmatrix}\);
- And lastly, we need to compute \((ps-rq)I_2=\begin{pmatrix} (ps-qr) & 0 \\ 0 & (ps-qr)\end{pmatrix}\);
- Putting it all together, the relationship is verified: \(X^2 - (p+s)X + (ps-rq)I_2=O_2\).

- Prove that if \(A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R})\) checks the relationship \(A^2=O_2\), then \(a+d=0\) and \(ad-bc=0\).
- If we look closely: \(a+d\) is the
*trace*of A \(tr(A)\); - Again, \(ad-bc\) is actually the determinant of \(A\);
- Now, if \(A=O_2\), then both the trace and determinant are \(0\);
- If \(A \neq O_2\), the relationship proved at the previous point becomes \(A^2 - tr(A)*A + det(A)*I_2=O_2\).
- But \(A^2=O_2\), so the relationship becomes: \(tr(A)*A=det(A)*I_2\).
- If we multiply each side of the previous equality with \(A\), and remember that \(A^2=O_2\), we get to \(det(A)*A=0\). But because we’ve decided that \(A \neq O_2\), we know that \(det(A)=0\), then \(ad-bc=0\).
- Given that \(det(A)=0\) and \(A^2=O_2\), then \(\underbrace{A^2}_{0}-tr(A)*A + \underbrace{det(A)*I_2}_{0}=O_2\), then \(tr(A)*A=O_2\), then \(tr(A)=0\), then \(a+d=0\).

- If we look closely: \(a+d\) is the
- Prove that if \(B \in M_2({\mathbb{R}})\) is
*nilpotent*, then \(B^2=O_2\).- If \(B\) is
*nilpotent*, then a number \(n \in \mathbb{N}^{*}\) exists, such that \(B^{n}=O_2\); - If \(B^n=O_2\), we can say the same about \((B^n)^2=O_2\).
- Given the previously proved relationship, we also know that \(tr(B^n)\) and \(det(B^n)\) are \(0\).
- As a known identity, \(det(M * N)=det(M)*det(N)\), \(∀ M, N \in M_m(\mathbb{C})\), so inductively we keep in mind the fact that \(det(M^k)=det(M)^k\);
- Now, let’s get back and find our \(n\). We introduce \(n_0\), \(n_0 \in \mathbb{N}^{*}\), the first \(n\) that satisfies the condition;
- We know that \(det(B^{n_0})=0\), and so \(det(B)^{n_0}=0\). The only possible way to satisfy that is if \(det(B)=0\).
- So the relationship proved at
`3.`

becomes \(B^2-tr(B)*B=0\). If we multiply each side with \(B^{n_0-2}\), we get \(B^n_0-tr(B)*B^{n_0-1}=O_2\). - But we already said that \(B^{n_0}=O_2\), this means that \(tr(B)*B^{n_0-1}=0\) so \(tr(B)=0\)
- At this point, all the conditions are met, and we can safely say that \(B^2=O_2\);

- If \(B\) is
- Prove that \(I_2\) cannot be written as an infinite sum of nilpotent matrices;
- Ad absurdum, we say that \(I_2=\sum_{k=1}^{n}*A_k\), where \(A_k\) is
*nilpotent*and \(k \in [1, \infty)\). - Because \(A_k\) is
*nilpotent*, then \(A_k^2=O_2\) and \(tr(A)=0\) (see`4.`

); - Through induction we can prove that \(tr(\sum_{k=1}^{\infty}A_k)=\sum_{k=1}^{\infty} tr(A_k)\);
- If our supposition that \(I_2=\sum_{k=1}^{n}*A_k\) was correct, then \(tr(\sum_{k=1}^{\infty}A_k)=0\). But that would be incorrect because \(tr(I_2)=2\), so our supposition was proved to be false.

- Ad absurdum, we say that \(I_2=\sum_{k=1}^{n}*A_k\), where \(A_k\) is

- Verify that \(\frac{1}{1-a}=1+a+...+a^n+\frac{a^{n+1}}{1-a}\), \(∀ n \in \mathbb{N}\) and \(∀ a \in \mathbb{R}-\{1\}\).
- We observe that \(1, a, a^2, a^3, ... , a^n\) are the first \(n+1\) terms of a
*geometric progression*that starts with \(1\) and has \(a\) as the ratio. - Using the formula, \(1+a+a^2+...+a^n=\frac{1-a^{n+1}}{1-a}=\frac{1}{1-a} - \frac{a^{n+1}}{1-a}\).
- Thus, \(\frac{1}{1-a}=1+a+...+a^n+\frac{a^{n+1}}{1-a}\)

- We observe that \(1, a, a^2, a^3, ... , a^n\) are the first \(n+1\) terms of a
- Prove the relationship \(\frac{1}{1+\sqrt{x}}=1-\sqrt{x}+(\sqrt{x})^2+...+(-1)^n(\sqrt{x})^n+(-1)^{n+1}\frac{(\sqrt{x})^{n+1}}{1+\sqrt{x}}\), \(∀ x \in [0,1], ∀ n \in \mathbb{N}\).
- Just by looking at
`1.`

and`2.`

we see a similar pattern. - The solution is to substitute \(a=-\sqrt{x}\), and the required relationship is proven.

- Just by looking at
- Prove that \(0 \leq \frac{(\sqrt{x})^{n+1}}{1+\sqrt{x}} \leq (\sqrt{x})^{n+1}\), \(∀ x \in [0,1], ∀ n \in \mathbb{N}\).
- We prove the two equalities one by one.
- We know that \((\sqrt{x})^{n+1} \geq 0\);
- We also know that \(\sqrt{x}+1 \geq 0\).
- We can then conclude that the ratio is also positive \(\frac{(\sqrt{x})^1} \geq 0\).
- \(\sqrt{x}\) is also an increasing function for \(x\) in our domain, so that the second is true;

- Prove that \(\lim_{n \rightarrow \infty}\int_{0}^{b}\frac{(\sqrt{x})^{n+1}}{\sqrt{x}+1}dx=0\), \(∀ \in [0,1]\).
- We make use of the previously proven inequality so we can
*squeeze*our integral like this: \(0 \leq \int_{0}^{b}\frac{(\sqrt{x})^{n+1}}{\sqrt{x}+1}dx \leq \int_{0}^{b}(\sqrt{x})^{n+1}dx\) - We determine \(\int_{0}^{b} (\sqrt{x})^{n+1}dx=\int_{0}^{b}x^{\frac{n+1}{2}}dx=\frac{x^{\frac{n+1}{2}+1}}{\frac{n+3}{2}}\Big\|_0^b=\frac{2}{n+3}*b^{\frac{n+3}{2}}\)
- We now use this in the limit: \(\lim_{n \rightarrow \infty}\int_{0}^{b} (\sqrt{x})^{n+1}dx=\lim_{n \rightarrow \infty} \frac{2}{n+3}*b^{\frac{n+3}{2}}\)
- We can write the limit \(\lim_{n \rightarrow \infty} \frac{2}{n+3}*b^{\frac{n+3}{2}}=(\lim_{n \rightarrow \infty}\frac{2}{n+3})*(lim_{n \rightarrow \infty}b^{\frac{n+3}{2}}) = 0 * k = 0\), where \(k \in \{0,1\}\).
- At this point, it is squeezed between two \(0\)s, so we’ve proven the limit to be 0;

- We make use of the previously proven inequality so we can
- Compute the following integral: \(\int_{0}^{b}\frac{1}{1+\sqrt{x}}dx\), \(b > 0\).
- We the following substitution: \(y=1+\sqrt{x}\), so that \((1+\sqrt x)dx=dy\), or \(2*\sqrt{x}dy=dx\), or \(dx=(2y-1)dy\).
- If \(x=0\), then \(y=1\)
- If \(x=b\), then \(y=1+\sqrt{b}\)
- So \(\int_{0}^{b}\frac{1}{1+\sqrt{x}}dx = 2 * \int_{1}^{1+\sqrt{b}} \frac{y-1}{y}dy=2*\int_{0}^{1+\sqrt{b}}dy - \int_{1}^{1+\sqrt{b}}\frac{1}{y}dy\);
- After every computation is done, the result is: \(\int_{0}^{b}\frac{1}{1+\sqrt{x}}dx=2*(\sqrt{b}-ln(1+\sqrt{b}))\)

- Prove that \(\lim_{n \rightarrow \infty} (x + \frac{(-1)^1x^{(\frac{1}{2}+1)}}{\frac{1}{2}+1} + \frac{(-1)^2x^{(\frac{2}{2}+1)}}{\frac{2}{2}+1} + ... + \frac{(-1)^nx^{(\frac{n}{2}+1)}}{\frac{n}{2}+1})=\int_{0}^{x}\frac{1}{1+\sqrt{t}}dt\), \(∀ x \in [0,1]\).
- We start with the integral. We’ve already worked on it at
`2.`

. - We change to \(t\), \(\int_{0}^{x}\frac{1}{1+\sqrt{t}}dt=\int_{0}^{x}(1-\sqrt{t}+(\sqrt{t})^2+...+(-1)^n(\sqrt{t})^n + (-1)^{n+1}\frac{(\sqrt{t})^{n+1}}{1+\sqrt{t}})dt\).
- This also can be written as a sum of two integrals: \(\int_{0}^{x}\frac{1}{1+\sqrt{t}}dt=\int_{0}^{x}(1-\sqrt{t}+(\sqrt{t})^2+...+(-1)^n(\sqrt{t})^n)dt+\int_{0}^{x}((-1)^{n+1}\frac{(\sqrt{t})^{n+1}}{1+\sqrt{t}})dt\)
- The relationship then becomes: \(\int_{0}^{x}\frac{1}{1+\sqrt{t}}dt=(x+\frac{(-1)*x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+...+\frac{(-1)^{n}*x^{\frac{n}{2}+1}}{\frac{n}{2}+1})+(-1)^{n+1}\int_{0}^{x}\frac{(\sqrt{t})^{n+1}}{1+\sqrt{t}}dt\)
- We apply the limit on the last relationship: \(lim_{n \rightarrow \infty}\int_{0}^{x}\frac{1}{1+\sqrt{t}}dt=\lim_{n \rightarrow \infty}(x+\frac{(-1)*x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+...+\frac{(-1)^{n}*x^{\frac{n}{2}+1}}{\frac{n}{2}+1})+\underbrace{lim_{n \rightarrow \infty}((-1)^{n+1}\int_{0}^{x}\frac{(\sqrt{t})^{n+1}}{1+\sqrt{t}})dt}_{0}\)
- Our relationship is proved.

- We start with the integral. We’ve already worked on it at

A good Baccalaureat grade was insufficient to get admitted to the UPB. You had to pass an additional admission exam. This one was a little harder than the *Bacaluareat*, but not impossible. The catch was not to make any mistakes. Any wrong answer could’ve been fatal.

On day one, you had to pass an exam composed of Algebra + Real Analysis, and on the second day, you had a Physics Exam where you were free to pick two areas. I’ve decided to pick *Classical Mechanics* and *Thermodynamics*.

In this section, I will cover only the Math part.

You had two hours to solve the following:

- Determine \(m \in \mathbb{R}\) so that \(f:\mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=\begin{cases}x^2-2x+m &, m \leq 1 \\ e^x-e &, x \gt 1\end{cases}\) to be continuos on \(\mathbb{R}\):
- Both \(x^2-2x+m\) and \(e^x-e\) are elementary functions, so we can say they are continuous on their intervals;
- So we need to see what’s happening when \(x=1\);
- We need to check the limits (from both sides) for \(f(x)\);
- The limit from the left: \(\lim_{x \rightarrow 1}f(x)=f(1)=m-1\)
- The limit from the right: \(\lim_{1 \leftarrow x}f(x)=f(1)=0\)
- The conclusion is simple \(m-1=0\), so \(m=1\)

- Solve the following inequation: \(\sqrt{x} \lt 1\).
- This is a simple one: \(x \in [0, 1)\)

- What’s the value for the following expression: \(E=\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}\)?
- The expression is equivalent to: \(E=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{3}+\sqrt{2}}{3-2}=2\sqrt{3}\)

- Compute the limit \(\lim_{n \rightarrow \infty}(\sqrt{n^2+n}-\sqrt{n^2+1})\).
- The limit is equivalent to: \(\lim_{n \rightarrow \infty}(\frac{n^2+n-n^2-1}{\sqrt{n^2+n}+\sqrt{n^2+1}})\);
- Then we do the following trick: \(\lim_{n \rightarrow \infty}(\frac{n(1-\frac{1}{n})}{n*(\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{1}{n^2}})})\)
- But \(\frac{1}{n} \rightarrow 0\) and \(\frac{1}{n^2} \rightarrow 0\) when \(n \rightarrow \infty\), so if we take that into the account the answer is \(\frac{1}{2}\).

- Given the equation \(x^2-ax+4=0\), \(a \in \mathbb{R}\). \(x_1, x_2\) solutions to the equation verify \(x_1 + x_2=5\) if:
- a) \(a=4\); b) \(a=0\); c) \(x_1=x_2\); d) \(a <0\); e) \(a=5\); f) \(x1, x2 \not\in \mathbb{R}\)
- Well, \(x_1+x_2=a\), then the correct answer is \(a=5\)

- The solutions for the equation \(9^x-4*3^x+3=0\) are?
- We consider \(y=3^x\), so the equation in \(y\) becomes: \(y^2-4y+3=0\);
- Then, \(y^2-3y-y+3=0\), or \((y-3)(y-1)=0\), the solutions are \(y_1=3\) and \(y_2=1\);
- This means \(3^x=3^1\), the exponential is injective, we can safely say \(x_1=1\) one of the two solutions;
- It also means \(3^x=3^0\), so \(x_2=0\) is the other solution

- Determine \(m, n \in \mathbb{R}\) so that the equation: \(x^4+3x^3+mx^2+nx-10=0\) to admit the solution \(x_1=i\):
- We substitute \(x=i\), so the equation becomes: \(i^4+3i^3+mi^2+ni-10=0\)
- We obtain \((-m-9) + i(n-3)=0\), this is a complex number.
- \((-m-9)\) needs to be \(0\), \(n-3\) needs to be \(0\)
- The answer is \(m=-9\) and \(n=3\)

- Compute the 10th term of the arithmetic progression starting with \(a_1=5\), with the \(r=2\).
- Using the formula we get: \(a_10=a_1+(10-1)*r=23\)

- Compute \(\int_{0}^{1}\frac{x^2}{x^3+1}dx\)
- We write as: \(\int_{0}^{1}\frac{x^2}{x^3+1}dx=\frac{1}{3}\int_{0}^{1}\frac{3*x^2}{x^3+1}dx\)
- We observe the integral can further be written as: \(\frac{1}{3}\int_{0}^{1}\frac{(x^3+1)'}{x^3+1}dx\)
- This is a known form, the answer is: \(\frac{1}{3}ln(2)\)

- If \((a,b)\) are a solution for the following system of equations: \(\begin{cases} x+y=2 \\ xy=1\end{cases}\), then:
- a) \(a^2b^2=2\); b) \(a^2+b^2=2\); c) \(a^2+b^2=1\); d) \(a^2+b^2=3\); e) \(a^2 + b^2 < 0\); f) \(a \neq b\)
- We simply solve the system, we see that \(a=1\), and \(b=1\), so the answer is \(a^2+b^2=1\)

- Given \(f:\mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=\frac{x^2}{x^2+1}\). Compute \(f'(1)\):
- We know \(f'(x)=\frac{2x(x^2+1)-x^2*2x}{(x^2+1)^2}\)
- We compute \(f'(1)=\frac{1}{2}\)

- The following composition rule is defined on \(\mathbb{R}\): \(x * y = xy+2ax+by\). Determine the relationship between \(a\) and \(b\) so that the rule is commutative.
- If the rule is commutative, then: \(xy+2ax+by=yx+2ay+bx\)
- After a few more steps: \(2a(x-y)-b(x-y)=0\), or \((x-y)(2a-b)=0\).
- The answer is: \(2a=b\) almost 20 years ago

- What’s the minimum of the following function: \(f:\mathbb{R} \rightarrow \mathbb{R}\), \(f(x)=\sqrt{4x^2+28x+85}+\sqrt{4x^2-28x+113}\)?
- This is pure grind, nothing exciting; the eventual answer will be: \(14\sqrt{2}\).

- Given \(f:\mathbb{C} \rightarrow \mathbb{C}\), \(f(z)=z^2+z+1\), compute \(f(\frac{-1+i\sqrt{3}}{2})\)
- We compute: \(f(\frac{-1+i\sqrt{3}}{2})=\frac{1-3-2*i\sqrt{3}}{4}+\frac{-2+2i\sqrt{3}}{4} + \frac{4}{4}\)
- The final answer is 0;

- Solve the following equation: \(det(\begin{pmatrix} 2 & x & 0 \\ x & -1 & x \\ 2 & -5 & -4\end{pmatrix})=0\)
- Computing the determinant, \(det=x^2-5x+4=0\), \(x=1\) or \(x=4\).

- Compute the limit of \(a_n=\sum_{k=1}^{n}(\frac{k(k+1)}{2x^{k-1}})\), where \(\vert x \vert > 1\)
- After all the grind the result is: \(\frac{x^3}{(x-1)^3}\)

- Given the function: \(f:[0, \infty)\), \(f(x)=\int_{x}^{x+1}\frac{t^2}{\sqrt{t^4+t^2+1}}dt\). Decide if:
- a) f(0) = 0; b) f is odd; c) f is convex; d) f admits a horizontal asymptote; e) f has 2 extreme points f) f admits one oblique asymptote
- We start checking for the following conditions:
- \(f(0)\) is false, f(x) is strictly \(>0\);
- \(f(x)\) cannot be odd, given its domain;
- so both a) and b) are false;

- We differentiate: \(f'(x)=\frac{(x+1)^2}{\sqrt{(x+1)^4 + (x+1)^2 + 1}} - \frac{x^2}{\sqrt{x^4+x^2+1}}=0\)
- After some grind, I am too lazy to write in Latex, we find out that there are no extreme points and and it doesn’t admit any horizontal asymptote
- The final result: \(f''(x)>0\) the the function is convex;

- Compute: \(\lim_{x \rightarrow 0}\frac{(x-1)^2-1}{x}\).
- After applying L’Hopital the result is: \(-2\)

Looking back at everything that was thrown at me 20 years ago, I have a few observations:

- Most of the exercises lack imagination and are either purely theoretical, like the one with matrices from the
*Bacalaureat*or extremely*grindy*; - The exams were not complex, but you had to get perfect scores under difficult time constraints;
- The system works, but it lacks any direction.
- In a follow-up article, I plan to select some Olympiad questions I have had to solve through the years. Those are much more interesting and imaginative.

That’s it. How were your exams?

]]>And then there are movies like the one David Cronenberg wrote and directed in 2022, called, unoriginally, *Crimes of the Future*. What a chilling unforgettable *abomination*! Imagine Lynch suddenly decides to make more sense, Bowie is in Berlin searching for the bluebirds, and the two hang together with Murakami and Koontz to write a novel and *stitch* some paintings together under the exuberant supervision of Ghenie. Everyone wants to shock the audience, but Cronenberg truly delivers. No, it’s not a masterpiece you potentially haven’t heard of, but this is not forgettable cinematography; I promise that. Some scenes are plain repugnant, and unsuitable for a sensitive or unprepared audience. Why would you kill “*Le vent nous portera*” in reverse Oedipal ways?

(Spoilers ahead: stop reading the article if you prefer watching the movie)

The movie is about the future, it is murder. Society is tired, people are comfortably numb, pleasure is rare, and nobody *feels* it anymore. Pain is the only mechanism that keeps everyone alive; humanity reduces to a flock/herd/swarm/colony/murder/exaltation/poverty of cenobites. The human body evolves at an accelerated rate, and new organs always appear, their functions unknown; no X-Men for today, sorry. Extirpating (or tattooing!) the *appendices* without anaesthesia in front of a live audience is the most popular form of entertainment society consumes. Degenerate individuals pay to see how “evolution is killed and mutilated” only to be regenerated a few moments later.

A State Ministry also deals with this newly extirpated genetic material. As you would expect, they are corrupt, and they lack digitalization. There will be no cloud providers in the future, and the biggest crime is the lack of smartphones; technology has turned into anarchy. What kind of future is that without touchscreens? At least the haptics are good, if not a little too sharp.

Meanwhile, in an eco-friendly twist, a secluded few have evolved organs to digest and nourish plastics.

That’s *eat*.

And then there are torture chairs and dining tables (?) that are extremely sought after. There’s a scene rivaling the one with Keanu Reaves versus Coppola’s brides (of Dracula), but there are no brides, only a vintage sarcophagus that cuts deep and then hastens the healing process of two level 2 support ladies. They are the twins from Kafka’s Judgement, but experts in mechanical scalpels and scissors.

That’s it; after watching this piece of cinematography, I couldn’t think for an hour straight, wondering myself, wtf did I just see? The plot is convex; it doesn’t hold water; everything is slow, gray, and oppressive by design. This is not Blade Runner or the rain falling on Rutger Hauer’s shadowing synthetic tears; but somehow all sits together, like a classic progressive rock album boring at times, with sparks of genius and virtuosity hidden in 20-minute long songs.

Should you watch this, know how immune you are to absurd (but well-deserved) acts of violence, torture, and Kristen Stewart, but let’s put it like this: I mostly blog about other stuff, and still you are reading this…

Cronenberg delivers with the ease of Nick Cave holding a rock in his hand.

(Did you know that eight cenobites are called a cenobyte?)

]]>How does a person answers

Why something happens?For example, Aunt Minnie is in a hospital, why?, because she slipped. She went out and she slipped on the ice and broke her hip. That satisfies the people, (…), but it wouldn’t satisfy someone who came from another planet, and knew nothing about it.

So first you need to understand why when you break a hip, you go to a hospital. How do you get to the hospital when the hip is broken? Well, because het husband seeing the hip is broken called the hospital up and sent someone to get her.

All that is understood by people.

Now, when you explain a

why?you need to be in a framework that would allow something to be true. Otherwise you are perpetually askingwhy?. (…)If you want to follow something up (n.a. by asking

why?) you go deeper and deeper in various directions.

The video continues to split in its beautiful directions; I recommend you to watch it in its fullness as an intellectual exercise.

And then I realized something every parent is aware of (or frustrated with): the endless chain of *whys* a young kid asks. Nothing stops them. They will continue digging *(why?-ing)* deeper and deeper until the only answer you can provide them doesn’t (even) satisfy you; it becomes impossible to address the question without losing in burdensome details or diverging from the original ask.

So, in a way, kids, by joining and assimilating into this world, are similar to Feynman’s aliens. They lack connection with the *meta-framework* we adults have inherited and are perpetuating, so they are constantly testing its malleable boundaries. They learn from us when to stop. For the mathematically inclined, remember the chain(ing) rule for computing annoyingly complex derivatives. As you get closer to the solution, a *residue* lurks to be calculated. So it is.

Raising a curious child is an exercise in becoming humble. From their honest questions, we can realize how much we don’t know and how many things we take for granted. For the kids, I suppose it’s not (even) curiosity but nature’s onboarding process in a strange new world waiting to be conquered and understood (maybe the two concepts are the same).

Parents with kids have a second chance to re-evaluate their *surroundings* through another pair of eyes, not their own, but of their own. Nature shows some mercy to the parents. It puts them in a position to share moments of exuberant curiosity with their offspring. It schools them so they don’t fail the class again.

And then I had a second epiphany; the real question was never *how?*. A limited set of answers for *how?* is already hardcoded in the biology of most animals. Animals (and even plants) know enough *how?* to survive and perpetuate. Look at them respectfully; we share the planet with some of the most ruthless and well-adapted biological organisms that ever lived (except for the panda bears).

Since our ancestors decided to look at the sky, the question that made us human in the first place was always *why?*. Revolutions start with *why?*, and then they are corrupted by people who only care about the *how?*.

If it is between one of the poor of your city and one of the poor of another city, the one of the poor of your city takes precedence. (Talmud, Baba Metzia 71a).

Don’t get me wrong; I don’t care much about what the so-called “sages” of old have said and how their opinions are radically different but valid simultaneously, like quantum “Talmudic” perspectives that sometimes collapse under their open claims. For a Universe, so ordered and mathematical, religious commentaries are notoriously speculative and relativistic. But what isn’t if there are humans around? This saying stuck with me, like some tool I’ve bought for fixing something in the house but never had the chance to use it properly. You remember the tool’s closeted existence, and you contemplate on justifying the Initial investment.

The year is 1970, my father is 11, my mother is 8, and Gil Scott-Heron releases a song called *Whitey on the Moon*.

(…)

A rat done bit my sister Nell

With whitey on the moon

Her face and arms began to swell

And whitey’s on the moon

I can’t pay no doctor bills

But whitey’s on the moon

(…)

How can anyone be so offended by this lovely scientific trip to the moon and back? It’s our shared victory; just like in sports, the human team scored against the vast, cold, extremely hot at times, and viciously inhospitable (team) Universe. It’s so empty; when it’s not, it wants to break your chemistry. But, we’ve just conquered Space by travelling one light-second to excellence, to the moon and back.

Thirty-five years later, another American musician, Darondo, released a brilliant album and a song called: *Let my people go*. Same topic, same lyrics, more music (in the classical sense) than spoken poetry:

Man, to your rocket ship

Take you to the moon

A million-dollar mission

To bring back a piece of rock

We got starvation, panic across the land

And here’s a fool in a rocket ship

Trying to be Superman

And then I realised that going to space is like feeding the poor of another city without taking care of your own. Which city?

But what if cities are within cities, and where do their borders end?

Furthermore, if you are a king ruling over a kingdom with multiple cities, which one is your city? The capital city? The Palace? Your vast bed chambers or the poor thoughts running through your head?

What if you live alone; people who live alone can’t build physical rocket ships. No offence, please, *people who live alone*.

What if all the people of your city are poor, but they don’t realise this fact about them? They are flawed in ways that are not obvious to an uneducated eye, and so are they: uneducated. But does it matter how educated your eye is if we’ve got starvation and panic across the land? Darondo asks.

Is it a sin to invest time and money in something so immediately useless as going to the moon without properly fixing your city first? Rocks are boring, the moon has almost no atmosphere, the view is not great, and the “climbing” (or “diving”) equipment is quite heavy even in low gravity.

Is this trip as “useless” as being a mathematician of the old centuries, spending unhealthy amounts of time studying prime numbers, while the poor are wearing wooden sandals and are dying of malnutrition, wars and pneumonia? After all, who needs your primes and zeta functions, when there is not enough bread on the tables or wheat in the granaries?

What if the poor are not poor? Whom should you help first?

What if you are poor in your city? In this case, by the Talmudic advice based on proximity, you’d better help yourself first.

I don’t have the answer, but it’s an exciting topic to meditate on. Should we science our way in directions so remote from eradicating poverty, or should we dedicate everything we have to fix our world, and the rest can wait? But why hurry?

At least Bowie’s spiders haven’t sold the world.

]]>I did this exercise on two popular platforms where I still have active accounts. One is the most extensive professional network, where people still debate the advantages of working from home versus returning to the office. The second is the platform our parents and early (as in older) millennials hold dear.

The results were underwhelming, but I wanted to scroll more; maybe something of interest would pop out from nowhere to comfort my sensation of void and emptiness. I was thinking about how my fingers are so accustomed to the inertia of scrolling down that they don’t even listen when asked to stop.

The wall on the professional network is almost sad. People generally love to take themselves too seriously in professional environments. This weakness is projected in how they act and talk. It’s like they identify themselves with their jobs too much, and they let themselves affected by this phenomenon.

Let me list my findings:

- Item 1: A recruiter in my local market is looking for a copywriter for a “well-known brand” I never heard of. Around 40 people liked the announcement; the majority of them were recruiters. Nothing to complain about; this is considered a typical encounter for a jobs site that has become a social media platform.
- Item 2: An influencer invites us to like, follow and subscribe to a company website offering “soft skills training”. The training is made of recorded sessions, costing around 60 dollars. I cannot comment on the quality of the materials, but common sense is telling me it’s not something groundbreaking.
- Item 3: A company I heard of is recruiting a Senior Java Developer. It’s a generic role. It’s been the same role since three years ago. It’s called passive recruiting.
- Item 4: A person from across the ocean (relative to where I live), who was laid off six months ago, is looking for another opportunity and asks for likes and shares so he can get better visibility to the companies who are currently recruiting. This is a challenging situation, and I feel sorry for him. I wish him luck; getting fired, and having unpaid bills on the table while your money deposits are dwindling, is a traumatic experience.
- Item 5: A guy who loves to post something every hour shared a picture with a book and a Michael Jordan quote: “If you quit once, it becomes a habit. Never quit!”. No, he is not working for a tobacco company. The quote is remarkably dull and very situational once taken out of context.
- Item 6: A guy posted a tutorial on running zookeeper. It’s a 10 minutes long YouTube video, with 33 views, where he executes the steps from the documentation. The post has one comment: “Thank you for posting!”
- Item 7: A person who identifies as a “Certified Financial Adviser” posts a stock photo (I am not exaggerating; the watermarks are there) and a short quote by James Frick about taking risks and risk management. I won’t reproduce the selection. It’s not worth the risk of losing your time.
- Item 8: A person from my extended network gracefully invites us to view his certified achievement from a cloud provider. It’s 2023; companies still pay money to certify their people on “volatile” technologies. Other than that, congratulations to him; that’s a challenging certification to get.
- Item 9: A “startup co-founder” is posting a rant about how the recruitment process is broken, and he still receives rejection emails years after applying. I don’t believe him, but 227 professionals liked his post. The comments section is messy. A self-described “Ex-Director” believes HR people are leaving their brains at home while working. Other users are posting laughing emoticons. The conclusion is that HR is bad.
- Item 9: A well-known security company is posting about the impending risk of computer viruses spreading inside your company’s infrastructure. They have revolutionary solutions to keep viruses at bay. I have yet to click further.
- Item 10: A Python Developer posts an interesting article about the novelties introduced by the newest python version. I had low expectations initially, but it was a good/opinionated read that kept me occupied for the next 10 minutes. I’ve added the blog feed to my RSS reader.
- Item 11: A programming influencer posted a 112-page document called “A very useful collection of 100+ java programs with detail.” It seems like a long read, but the code needs to be better written and formatted. One hundred seventy-one people appreciate the coding ratatouille. The comments praise the author for his dedication, and the author responds with heart emoticons.
- Item 12: Another person in the business of A.I. (judging by the company he is working for) assures us that “high-status” professionals will lose the battle with A.I., and the future belongs to the amateurs that know how to navigate the new reality.
- Item 13: A company invites me to: “Unlock your IoT potential with our distributed intelligence developer program.” I like how they combine a decade-old buzzword (IoT) with a contemporary one, but discretely.
- Item 14: A meme about the imposter syndrome and the redesign of the Google Chrome logo. Unfortunately, I didn’t find it funny, but it attracted a generous amount of attention (a few hundred likes) and no comments.
- Item 15: A random but aesthetically pleasing picture of blossoming cherry trees accompanied by a heart emoticon. Because of the heavy filtering and processing, it doesn’t look real.
- Item 16: A data science influencer posts pictures with general information about gradient boosting. No mathematical formula is mentioned, but the text is hefty in scientific jargon. It gives me the impression the author needs to learn about what he is speaking about. He repeats terms like “gradients” and “weighted loss function” without making sense. But I am no expert.
- Item 17: A self-described “The Simplifier” encourages us not to mistake simplicity for the lack of depth. He concludes that complexity can be impressive, but
*simplicity*it’s the one that saves the day and the hidden place where the true genius lies dormant. At the moment I am writing this, there are no comments and 30 likes. Did I tell you love (good) complexity? - Item 18: A self-described “Innovation Leader” encourages us to be careful with whom we share our dreams because some might mistake dreams for arrogance. I was slightly disappointed he didn’t quote Eurythmics, the band. The post doesn’t resonate with me. So I tenderly ignore it.
- Item 18: An extremely popular podcaster asks us some good questions to ask his next guest. Three hundred people have commented. Some of the questions are very good. This podcaster is known for his work on A.I. and for interviewing people I admire Knuth, Stroustrup, and Van Rossum.
- Item 19: A person shared with us that she loves dark chocolate, and a Harvard study found heavy metals in popular brands of dark chocolate.
- Item 20: A job ad for an Android Developer. The title is unfortunate: “Are you an Android Developer who likes to live on the edge? It could be a call for help or a subliminal warning. Living on the edge, it’s not something the average developer aspires to, especially in those times and days.
- Item 21: A software engineer writes about the cases when O(n) is better than O(logn). I am not convinced, but it’s something I would think of.
- Item 22: A UI/UX designer shares that he is in the top 10% of designers worldwide regarding attention to detail. There is an online test for this.
- Item 23: A person I’ve worked with is changing jobs. Good decision. I congratulated her.
- Item 24: A junior developer is looking for an internship or an entry-level job. The message is wholesome. Unfortunately, she classified react, HTML, and CSS as programming languages.
- Item 25: Someone thanked one of his colleagues for going above and beyond (Kudos!).
- Item 26: A recruiter is looking for SAP Hybris experts to join an ongoing project. Remote work is accepted.
- Items 26: A self-described “Communication Strategist” shared an article about how to make your employees feel appreciated. One of the suggestions is to give them shout-outs on social media. This might explain the kudos from Item 25.
- Item 27: A programming influencer explains (in pictures) why
*multiple inheritance*is bad and it doesn’t work in Java. - Item 28: A Software Developer with more than 20 years of experience is announcing to his network that he decided to quit his job and go full-time freelancer. He invites people to contact him for contracting work.
- Item 29: An ex-colleague of mine changed jobs.
- Item 30: A crypto influencer tells us the 2024 A.I. bubble will be insane.

And here I’ve stopped “analysing” my personalised wall. The algorithm failed to display valuable content. It’s *Static Noise*, with modest spikes of statistical relevance.
So I’ve decided to quit mainstream social media sites. This time for good. Hopefully. Time is a valuable resource. Life is short. Social Media feeds on our time.

Social media, I quit!

]]>This post is satire. It may also not make too much sense.

Today I’ve seen this picture on my LinkedIn wall:

And it triggered me badly. There are people in my network highly sensitive to the *divine nature* of numbers. As a math enthusiast, I won’t get it.

So, for all the people believing in numerology, let’s find the numbers that carry highly amplified or/and uber-vibrational powers, more than you can imagine.

Keep in mind `37`

is super-weak compared to `15873`

:

```
The magic of the number 15873
15873 x 7 = 111111
15873 x 14 = 222222
15873 x 21 = 333333
15873 x 28 = 444444
15873 x 35 = 555555
15873 x 42 = 666666
15873 x 49 = 777777
15873 x 56 = 888888
15873 x 63 = 999999
(As you can see, 15873 times a multiple of 7, gives a sextuple Rep Digit,
short for Repeating Digit which carry amplified or vibrational powers.
```

But let me tell you, even `15873`

is super-weak compared to other numbers that carry *uber-highly-amplified and/or uber-highly-vibrational powers*. So let’s write a small python script that can list more *vibrational powerful numbers* (whatever that is) than `37`

and `15873`

.

```
from primefac import primefac
from collections import Counter
import sys
def compute_vibrational_num(vp):
n, res, vp_ret = 1, 0, vp
while vp != 0: n, vp = 1000 * n + 1, vp - 1
n *= 111
primef = Counter(list(primefac(n))).most_common(2)
primef = primef[1 if primef[1][0] == 7 else 0]
magic = primef[0] ** primef[1]
res=n//magic
return (magic, int(res), vp_ret)
def print_text(tpl):
print(f"The magic of the number {tpl[1]}")
for i in range(1,10):
print(f"{tpl[1]} x {tpl[0]*i} = {tpl[1] * tpl[0] * i}")
print(f"""
(As you can see, {tpl[1]} times a multiple of {tpl[0]}, gives a {(tpl[2]+1)*3}nthuple Rep Digit,
short for Repeating Digit which carry amplified or vibrational power.
""")
```

Yes, I know, it looks bad, and you also have to install a pip package called primefac to do the *hard work* for us. Now, the `vp`

parameter from our function gives us the *vibrational power* (whatever that is) of the number.

For example, if we call `print_text(compute_vibrational_num(0))`

, we get the results from the picture (that’s boring):

```
The magic of the number 37
37 x 3 = 111
37 x 6 = 222
37 x 9 = 333
37 x 12 = 444
37 x 15 = 555
37 x 18 = 666
37 x 21 = 777
37 x 24 = 888
37 x 27 = 999
(As you can see, 37 times a multiple of 3, gives a 3nthuple Rep Digit,
short for Repeating Digit which carry amplified or vibrational power.
```

But if we increase the vibrational power of the number to, let’s say, `vp=13`

, we will get something significantly more powerful:

```
The magic of the number 2267573696145124716553287981859410430839
2267573696145124716553287981859410430839 x 49 = 111111111111111111111111111111111111111111
2267573696145124716553287981859410430839 x 98 = 222222222222222222222222222222222222222222
2267573696145124716553287981859410430839 x 147 = 333333333333333333333333333333333333333333
2267573696145124716553287981859410430839 x 196 = 444444444444444444444444444444444444444444
2267573696145124716553287981859410430839 x 245 = 555555555555555555555555555555555555555555
2267573696145124716553287981859410430839 x 294 = 666666666666666666666666666666666666666666
2267573696145124716553287981859410430839 x 343 = 777777777777777777777777777777777777777777
2267573696145124716553287981859410430839 x 392 = 888888888888888888888888888888888888888888
2267573696145124716553287981859410430839 x 441 = 999999999999999999999999999999999999999999
(As you can see, 2267573696145124716553287981859410430839 times a multiple of 49, gives a 42nthuple Rep Digit,
short for Repeating Digit which carry amplified or vibrational power.
```

*I wish you to find the most vibrational number your hardware allows to.*

Repdigits are natural numbers composed of instances of the same digit. The coolest repdigits are the Mersene primes. They are prime numbers that, when represented in binary, are composed only of `1`

s.

All in all, let’s get back to our powerful *vibrational numbers* (whatever that means).

Numbers like `11...1`

can sometimes be prime, but if the number of `1`

digits is a multiple of 3, we know they aren’t. It’s a simple proof:

If we group the digits `3`

by `3`

, we’ll get the following relationship:s

And then if we use `111`

as a common factor we obtain:

But \(111=3 * 37\), so repdigits with the number of digits of \(3\), are always divisible with \(3\) or \(37\). As an interesting observation, they are from time to time divisible with \(7\) (but not always).

This is how our code functions:

- It generates numbers like
`111..111`

; - It gets their prime factors (we know for certain they are not prime);
- Then it separates the prime factors in a “bigger”
*highly vibrational number*, and keeps a multiple of`3`

or`7`

as the*meme*multiplier;