I’m so glad to see you here.

Three puzzles, three sips of math: Are you ready for your mid-December brain brew?

For today’s edition, I selected three puzzles from logic, geometry and probability. As usual, you’ll find the estimated difficulty marked with coffee cups ☕, a hint of how many cups you might need to tackle the problem (just a warning, not a recommendation!).

Note:

All puzzles 1 and 2 call for a touch of creativity.

🧩 The Puzzles

Puzzle 1 (Logic) — ☕☕☕

You’re in a room with three switches (A, B, C) that control three bulbs in the next room. You can flip switches any way you want, then go into the other room once to check bulbs. How do you identify which switch controls which bulb?

Hint: Use heat as well as on/off.

Puzzle 2 (Geometry) — ☕☕☕

Folding an A4 piece of paper yields a polygon (a geometrical figure) with N edges. The value of N depends on how you fold it. What are all the possible values for N?

Hint: Take a piece of paper and try it yourself!

Puzzle 3 (Probability) — ☕☕☕☕

100 passengers board a plane with 100 assigned seats.

• Passenger 1 has lost their boarding pass and sits in a random seat.

• Each subsequent passenger:

  • Sits in their own assigned seat if it’s available

  • Otherwise, picks a random empty seat

What is the probability that Passenger 100 (the last to board) ends up in their own assigned seat?

Hint: Don’t try to track all 100 passengers. Instead, ask yourself: Which seats actually “matter” to the final outcome? Think about what conditions would guarantee Passenger 100 gets their seat versus guarantee they don’t, and when those conditions get decided.

🔹 Suggested approach: Try the puzzles in order of difficulty, and, most importantly, have fun!

That’s all for this week’s brew of puzzles! How did you find them: Too easy, too tricky, or just right? I’d love to hear from you!

🗓️ Next week teaser

Next Tuesday, The Cafe will present a freshly brewed selection of AI prompts (my personal favourites).

I look forward to seeing you then.

Until then, have a wonderful week!

With love and curiosity,

Sybille ☕

P.S.: Solutions will appear in next week’s edition of The Café.

In the meantime, paid subscribers are very welcome to share their solving process and tentative answers using this form: I’ll send back personal feedback, tips, and good coffee vibes ☕💡.

Here’s the catch: In a room of just 23 people, there’s a 50% chance that two people share the same birthday. With 70 people, that probability jumps to 99.9%.

Oh, sure.

Wait … What?

There are 365 days in a year, so shouldn’t you need way more people?

Why Our Intuition Fails Us

We instinctively think, “What are the odds someone shares my birthday?” And you’d be right—with 23 people, the chance someone matches your specific birthday is only about 6%.

But the birthday paradox isn’t asking that. It’s asking: “What are the odds that any two people in the room share a birthday?” That’s a very different question, because you’re not comparing to just yourself—you’re comparing everyone to everyone else. This seemingly small shift in perspective changes everything.

The Power of Pairs

Here’s where things get interesting. When you’re comparing everyone to everyone else, you’re not making 23 comparisons—you’re making way, way more.

Let’s count them:

• Person 1 can be paired with 22 other people

• Person 2 can be paired with 21 remaining people (we already counted their pairing with Person 1)

• Person 3 can be paired with 20 remaining people

• And so on...

Add it up: 22 + 21 + 20 + ... + 2 + 1 = 22*23/2=253 different pairs

With 253 opportunities for a match across 365 possible days, suddenly a 50% probability doesn’t seem so outrageous, does it?

This is the key insight: the number of comparisons grows much faster than the number of people. The mathematical formula is n(n-1)/2, where n is the number of people. That’s what’s called “quadratic growth”; it accelerates rapidly.

• 10 people = 45 pairs

• 20 people = 190 pairs

• 30 people = 435 pairs

• 50 people = 1,225 pairs

• 100 people = 4,950 pairs

The Math: Why 23 is the Magic Number

Let’s work through the actual calculation. Fair warning: we’re going to get a bit mathematical here, but I promise to make it as painless as possible.

The trick is to calculate the opposite: Instead of finding the probability that two people do share a birthday, we’ll find the probability that nobody shares a birthday. Then we’ll subtract from 100%.

Why? Because calculating “nobody shares” is much simpler than calculating all the different ways people could share birthdays (two people share one date, three people share another date, two pairs share different dates, etc.).

Let’s Build It Step by Step

Person 1 enters the room. They can have any birthday. Probability that their birthday is “unique so far” = 365/365 = 100%

Person 2 enters. For nobody to share yet, Person 2 must have a different birthday than Person 1. There are 364 days available out of 365 total days. Probability = 364/365 = 99.73%

Person 3 enters. They must avoid both previous birthdays. Probability = 363/365 = 99.45%

Person 4: 362/365 = 99.18%

And we continue this pattern...

The probability that ALL people have different birthdays is:
(365/365) × (364/365) × (363/365) × (362/365) × ... and so on

For 23 people, this multiplication gives us approximately 0.493, or 49.3%

This means there’s a 49.3% chance that all 23 people have different birthdays.

Therefore, the chance that at least two people share a birthday is:
1 - 0.493 = 0.507, or 50.7%

The Complete Table

Here’s how the probabilities grow:

• 10 people: 11.7% chance of a match

• 15 people: 25.3% chance

• 20 people: 41.1% chance

• 23 people: 50.7% chance ✨

• 30 people: 70.6% chance

• 40 people: 89.1% chance

• 50 people: 97.0% chance

• 60 people: 99.4% chance

• 70 people: 99.9% chance

By the time you have 100 people, the probability exceeds 99.99997%. It’s essentially certain.

An Eerie Birthday Factoid

Remarkably, three of the first five U.S. Presidents died on July 4th (John Adams, Thomas Jefferson, and James Monroe).

Coincidence? Probably!

The Birthday Paradox is here to remind us that coincidences are more common than we think.

Why This Feels Like Cheating (But Isn’t)

Even after seeing the math, many people feel like some details have been missed, that there should be a mistake somewhere. Let me address the common objections:

“But what about leap years?” Including February 29th actually makes matches slightly more likely because there are fewer options. The paradox still works.

“What if birthdays aren’t evenly distributed?” Great catch! Birthdays actually aren’t evenly distributed—more babies are born in late summer/early fall in many countries. This increases the chances of a match, making the paradox even stronger.

“You’re not accounting for twins!” If we include twins, the probability of matches goes up even more. The 23-person threshold is actually a conservative estimate.

“This still feels wrong.” I know! I feel the same. That’s what makes it a paradox. Our intuition isn’t broken: it’s just calibrated for individual probabilities (what are the odds for me?) rather than collective probabilities (what are the odds for any pair in the group?).

Why It Actually Matters

Beyond being a great mind-experiment to sharpen your thinking, the Birthday Paradox has several practical implications:

Cryptography and Security

Hash functions (which are crucial for digital security) are vulnerable to “birthday attacks.” If a hacker needs to find two different inputs that produce the same output, the Birthday Paradox tells us they’ll succeed much faster than you’d intuitively expect. This is why security protocols need much larger hash spaces than you might think.

Data Deduplication

Database engineers use birthday paradox math to estimate how likely duplicate entries are in large datasets. This affects everything from social media platforms to financial systems.

DNA Testing

With millions of people using ancestry DNA tests, how many false matches might occur by chance? The Birthday Paradox helps scientists calculate these probabilities and set appropriate thresholds for “matches.”

Network Analysis

In social networks or disease transmission modelling, the Birthday Paradox principle helps predict how quickly connections (or infections) will spread through a population.

Quality Control

Manufacturing systems use this math to predict the likelihood of duplicate serial numbers, identical defects, or other collision events in large-scale production.

The Bigger Lesson

The Birthday Paradox is teaching us something profound about how we naturally think about probability:

We’re wired to think about individual probabilities, not collective ones. We naturally ask “what are MY odds?” rather than “what are THE odds?”. We view ourselves as the protagonist, not side character.

This bias affects how we think about risk, coincidence, and randomness in daily life. We’re shocked when unlikely things happen, but we forget to account for all the opportunities they have to occur.

Consider:

• “What are the odds I’d win the lottery?” Very low. “What are the odds someone wins in my city?” Pretty high, given millions of tickets.

• “What are the odds I’d run into someone I know at the airport?” Low. “What are the odds anyone at this airport runs into someone they know?” Much higher.

• “What are the odds I’d have a bizarre dream featuring dinosaurs transforming into butterflies last night?” Small. “What are the odds someone on Earth had such a bizarre dream?” Essentially 100% (maybe a bit lower for this one — still, you get the picture).

The Birthday Paradox is the perfect invitation to correct this bias. Math often reveals truths our intuition misses, and that the universe is both more predictable and more surprising than we may initially think.

The Bottom Line

The Birthday Paradox teaches us humility, reminding us that:

• Coincidences are more common than they feel

• The number of possible interactions grows faster than we realize

• Sometimes the counterintuitive answer is the correct one

• Math can reveal truths that hide in plain sight

So next time you’re at a party and two people share a birthday, remember: the real surprise isn’t that it happened—it’s that you thought it wouldn’t.

Next week

Next Tuesday, check your mailbox for three puzzles to challenge your mind.

Until then, I wish you a wonderful week 🌸.

With love and curiosity,

Sybille

I hope you enjoyed this post!

What’s your most surprising birthday coincidence story? Hit reply and tell me; I read every response.

No matter how complicated a map is, you never need more than four colours to colour it so that no two neighbouring regions share the same colour.

This is the essence of the Four Colour Theorem, one of the most famous results in topology and graph theory. At first glance it sounds like a cute puzzle. In reality, it turned out to be one of the hardest problems of the 20th century — and no one succeeded in solving it until computer-assisted proofs entered the game.

What this problem says … And doesn’t say

What it says

Take any map; delimitations (boundaries) on this map define regions. Each region is adjacent to number of other regions, meaning that they share a common boundary that is not a point; two regions that share isolated boundary points are not considered as adjacent.

Then the Four Colour Theorem states:

Any map can be coloured with at most four colours, so that no two adjacent regions share the same colour.

For the more mathematically-minded around you:

Here boundaries on the map are assumed to be finite unions of differentiable curves (‘real-life’ boundaries in the real plane); we forbid ‘regions’ delimitated by esoteric boundaries such as topological curves with infinite length. In particular, if you associate to each region of the map a vertex, and connect two vertices if the associated regions are adjacent, the Four Colour Theorem states that you can colour the vertices of the associated graph with only four colour, so that no two adjacent vertices receive the same colour.

What the problem doesn’t say.

It doesn’t state that you can colour country maps with 4 colours. Indeed, the cartography convention for colouring countries is that you always colour two regions belonging to the same country in the same colour even if the region is not adjacent to the ‘main’ part of the country.

And the four colour theorem isn’t true in that setting! You’ll usually need more than 4 colours to colour a ‘country map’. For example France and Netherlands share a boundary (in the Saint Martin island), which implies that countries in Europe need to be coloured with at least five different colours.

Some examples

Let us go to some examples together.

First, let us consider this pie:

Notice you can colour it using only three colours! (Remember that regions that share only points aren’t considered adjacent, so only consecutive ‘slices of the pie’ are considered adjacent)

Now let us consider the map of the USA. One possible colouring with four colours is:

So we can colour the USA map using only four colours.

Let us think a bit more…

Can we use less than four colours?

The answer is no. You can prove it by considering the regions surrounding the Nevada state:

• Let us pick a colour for the Nevada state (yellow in the map under).

• Then necessarily all five neighbouring states are coloured in colours other than yellow.

• We need three different colours at least to colour them (if you go through all the possibilities for two colours you always end up colouring two adjacent states with the same colour — which is a colouring style we have forbidden).

Your turn!

Difficulty: ☕☕

The sketch/map under has been coloured using five colours. Can you colour it using four colours only?

(Scroll down for the solution…)

A hundred-year history

A problem from the age of hand-drawn maps

Back in 1852, a British student named Francis Guthrie noticed while colouring a map of counties that four colours always seemed to be enough. He asked his brother, who asked mathematician Augustus De Morgan, who asked the wider mathematical community… and nobody could prove it.

For over a century, mathematicians tried—and failed—to show that there exists no map requiring five colours.

The surprising twist

After decades of partial progress, the breakthrough came in 1976 from Kenneth Appel and Wolfgang Haken.

They used a mix of old ideas along with the the new computing power of computers: instead of proving the theorem “by hand,” they reduced the problem to a finite but enormous set of special cases and checked them using a computer. If no counterexample exists among those cases, then no counterexample exists at all.

It was the first major theorem proven with essential computer assistance—causing both excitement and controversy. Some mathematicians were uncomfortable: could a proof be trusted if no human could check every line?

Today, the Four Colour Theorem is widely accepted, and later work has simplified and reverified the computational parts. But the debate about what counts as a “proof” still echoes in modern discussions about machine-verified mathematics.

What about human proofs?

Several mathematicians have tried providing a 100% human-written proof — with little success so far. Thomassen provided a beautiful proof in the 90s that every map can be colour with five colours only, but the four colour case remains in the realm of computers. This reminds us of how crucial computer assistance can be even in the most theoretical areas of science.

What makes this theorem beautiful

Beyond the historical drama, the theorem touches on deep ideas.

Maps are really just graphs in disguise. Regions become nodes; shared borders become edges. The theorem says that planar graphs (graphs that can be drawn without edges crossing) are four-colourable. It pushed forward the development of graph colouring, a very active field of research in graph theory.

It has applications in many practical areas (you probably use it every day), including:

• Scheduling and timetabling tools (softwares ensuring there are no conflicts in time slots/ room booking);

• Registering allocation in compilers (used for example by your web browser compiler, every day billions of people use these tools without knowing about it);

• Cloud resource allocation;

• Computer graphics/mesh colouring.

In the meantime, the proof forced mathematicians to invent new combinatorial tools and pushed the development of computational proof verification.

And yet—despite all the complexity—its statement remains disarmingly simple.

You can explain it to a child with a pack of coloured pencils (… and a lot of patience).

The Four Colour Theorem is a reminder that even problems that look naïve or recreational can open the door to deep mathematics. Some of the most profound discoveries start with curiosity, a simple question, or even… a map someone was trying to colour neatly.

📅 Next week

Next Tuesday, we will discuss a nice paradox involving your birthday date.

Until then, I wish you a beautiful, blessed week 🌸.

With love and curiosity,

Sybille

Solution

In recent months, countless people have been unfairly accused of AI use while actual AI users operate completely under the radar. This made me curious — and willing to investigate the phenomenon further.

Today I share with you what I investigated. Here’s what I found, in a word: AI detectors are dangerously unreliable, and are inducing us in error.

(Scroll down for the complete data I obtained)

What Are AI Detectors, Anyway?

Several tools claim to identify what percentage of your text was AI-generated. They highlight “suspicious” sections and promise to distinguish human writing from LLM output.

The premise sounds reasonable. The execution? Not so much.

My Test: What Actually Happens

I compared:

• 100% AI-written text (Claude Haiku 4.5 and Claude Sonnet 4.5)

• Professional human writing (opening chapters from Rich Dad Poor Dad and The Eye of the World)

• Non-professional human writing (my own math notes—no AI involved)

The detectors I tested:

• Pixozone

• Scribbr

• ZeroGPT

My hypothesis: professional writers would be flagged as human; amateur writers (like me) might look suspicious.

The AI Writing Prompt

Here’s the prompt I used with Claude Sonnet and Claude Haiku (version 4.5):

“Write a creative piece about a character meeting someone at a coffee shop in Mumbai. Channel your inner messy human writer:

- Start with a sensory detail that feels oddly specific
- Use at least one metaphor that’s slightly weird or imperfect
- Vary your rhythm: punchy fragments mixed with long, winding sentences
- Include moments where the narrator gets distracted
- Use unexpected adjectives (”aggressive sunset” not “beautiful sunset”)
- Drop in pop culture references or time period markers
- Let one sentence be slightly awkward or grammatically questionable
- End on something unresolved or ambiguous
- Write like you’re journaling at 2am—messy, with interruptions and trailing thoughts”

Why this prompt matters: I specifically asked Claude to write like a human with flaws, distractions, and imperfect phrasing. This is key to understanding why detectors fail.

Here’s part of What Claude Generated:

There was an uncle at the corner table eating a croissant with a fork and knife.

I watched him for probably thirty seconds, this meticulous dissection of pastry, and I wanted to tell him that’s not how you eat a croissant, you’re supposed to let it fall apart, let the flakes get everywhere, let it be messy, but then again maybe he had the right idea, maybe controlling the chaos was better than—

The door opened.

My entire body knew it was him before my brain caught up, this sudden awareness of my own pulse in places I didn’t know had pulse points—my wrists, my neck, that spot right behind my knee that was suddenly, inexplicably, sweating.

Why This Extract Matters

Notice the Claude text has:

✓ Specific, odd details (sitting at a corner table, 30 seconds)

✓ Imperfect metaphors (”my knee that was suddenly, inexplicably, sweating”)

✓ Awkward phrasing

✓ Stream-of-consciousness distractions

✓ Unresolved ending

It reads human because I asked it to.

This is exactly why detectors fail—they can’t distinguish between AI mimicking human messiness and actual human writing.

🚨 The Results: The Detectors Got It Backwards

What was tested/ How it was flagged

The numbers indicate which percentage of the text (on average) is estimated as being AI for each AI detector.

The clearest evidence:

Scribbr flagged ALL Claude-generated texts as ‘100% Human-written’, while ZeroGPT flagged a Claude-generated text as 0% AI (scroll down for a full breakdown of the data).

Meanwhile, some of my entirely human-written notes were labeled as AI by two of the detectors.

The Takeaway: Don’t Trust AI Detectors

These tools are unreliable, inconsistent, and potentially harmful.

They’re more likely to flag genuine human work than catch actual AI use.

:

:

:

:

Just don’t trust them.

How to Protect Yourself

If you’re a student or professional concerned about AI accusations:

✓ Be transparent. Clearly state which parts are 100% human-produced. For example: “The ideas and research are my own. I used AI to organize and refine the writing.”

✓ Keep your workflow visible. Archive drafts, edits, and earlier versions to show your creative process.

✓ Build a portfolio. Consistent work over time demonstrates your unique voice in ways detectors never can.

Bottom line: Rather than fear false accusations, focus on being honest about your process. Your track record and transparency will matter far more than what any AI detector claims.

Next week

See you next Tuesday for an excursion into the realm of map colouring — a problem that stumped mathematicians for a long time, until a strategic partnership between humans and computer programs yielded a neat proof!

Until then, I wish you a very beautiful week.

With love and curiosity,

Sybille ☕

The Data

You will find here three puzzles, suitable for advanced high school and university students.

As always, the difficulty is noted with coffee cups — the more cups the harder the problem.

Good luck!

🧩 Problem 1: Logic/ Algebra

Difficulty: ☕☕

Three children’s ages are whole numbers >0. Their product is 36. A neighbour knows their sum but still can’t determine the ages. The neighbour then learns “the youngest child likes chocolate ice cream” (i.e., there is a youngest child), and then figures the ages. What are the ages?

Hint: List triple factors of 36 and their sums; find ambiguity.

🧩 Problem 2: Probability/ Combinatorics

Difficulty: ☕☕☕

An absent-minded assistant placed three letters randomly into three envelopes. What is the probability that at least one of the recipients gets his letter ?

Hint. Count how many arrangements give no one their own letter (this is easier). Then use the complement:
Pr⁡(at least one gets correct)=1−Pr⁡(none get correct).

🧩 Problem 3: Algebra

Difficulty: ☕☕☕

Solve for real x:

Hint: Let the infinite radical equal y and form an equation for x; then translate back to x.

Stay tuned for more puzzles like these ones!

With love and curiosity,

Sybille ☕

Solutions

Solution of Problem 1

Answer: 1, 6 and 6 (the eldest are twins).

Triples whose product = 36:

• (1,1,36) → sum = 1+1+36 = 38.

• (1,2,18) → sum = 1+2+18 = 21.

• (1,3,12) → sum = 1+3+12 = 16.

• (1,4,9) → sum = 1+4+9 = 14.

• (1,6,6) → sum = 1+6+6 = 13.

• (2,2,9) → sum = 2+2+9 = 13.

• (2,3,6) → sum = 2+3+6 = 11.

• (3,3,4) → sum = 3+3+4 = 10.

The neighbour knows the sum but still can’t determine ages → the sum must be 13 (ambiguous between (1,6,6) and (2,2,9)). Then the neighbour learns “there is a youngest child” (i.e., not all same youngest age — distinguishes a unique younger child), which eliminates (2,2,9) because that has two youngest twins (2 and 2). So the ages are 1, 6, and 6.

Solution of Problem 2

The answer is: 2/3.

Number of total arrangements: 6 (3 choices for where to put letter A, 2 choices for which envelope for letter B, and only one choice remaining for which envelope for letter C).

Number of arrangements where no one gets their own letter: 2 (2 choices for letter A, and only one remaining choice for letters B and C).

Hence:

P(no one gets their own letter)=2/6=1/3

Finally:

P(at least one gets its own letter) = 1 - 1/3 = 2/3.

Solution of Problem 3

Answer: x=6.

Dear math friend,
I’m so glad to see you here!

For this Café edition, we’ll travel together through a story that spans thousands of years —
The story of a theorem you recently proved in our puzzles: the Pythagorean Theorem!

🏺 The Story: From Ancient Tablets to Greek Temples

You probably learned the theorem back in junior high school and didn’t think much about it:

In a right triangle, the square of the largest side — the hypotenuse, the side opposite the right angle — is equal to the sum of the squares of the two other sides.

You may even have learned a proof… and soon forgotten it (I know I did 😅).
But this familiar statement actually played an extraordinary role in the history of mathematics!

The first traces of the theorem go back more than 4,000 years.
A clay tablet from the First Babylonian Empire (c. 1900 B.C.E.), now known as Plimpton 322 and kept at Columbia University, lists columns of numbers that we now call Pythagorean triples: sets of integers (a, b, c) satisfying

a²+b²=c².

This strongly suggests that Babylonian scholars already knew at least some version of the theorem — not surprising given its role in astronomy, and the Babylonians’ advanced knowledge in the field.

In ancient India, the Baudhāyana Śulba-sūtra (circa 800–400 B.C.E.) states the theorem explicitly in special cases.

Meanwhile, ancient Egyptians regarded the 3−4−5 triangle as sacred: the hypotenuse (of length 5) represented the god Horus, and the other two sides (of lengths 3 and 4) gods Isis and Osiris (as described here by Greek historian Plutarch).

Centuries later, the theorem became central to the mystical school founded by the Greek philosopher Pythagoras (6th century b.c).
Legend has it that, upon proving the result, Pythagoras sacrificed a hundred oxen to the gods in gratitude, though modern historians suspect he may have learned it during his travels (for example in Egypt) and shared it with his Greek contemporaries, who were unaware of the earlier discoveries.

Nowadays, far from the mysticism of olden time, the theorem is a standard component of most junior high schools/ high schools curriculums around the world — with around 50 million students learning the theorem each year! Its generalizations play an important role in many fields, including civil and structural engineering, game design (normalizing movement speed, projectile aiming, collision detection, physics and camera calculations), GPS & satellites positioning (using spherical analogues of the theorem to determine distance formulas), and medical imaging (reconstructing 3D structures from 2D using MRI, CT or ultrasound).

✏️ Your Turn!

Now it’s time to challenge yourself!

Text within this block will maintain its original spacing when published

Below are two ancient diagrams. Each one hides a proof of a result related to the Pythagorean theorem. 
Can you spot it?

(—> Scroll down for the solutions!)

1️⃣ Zhoubi Suanjing’s diagram (China, ~100 B.C.E.)

Puzzle difficulty: ☕️☕️☕️

Hint: This diagram hides a geometric proof that the hypothenuse of a right triangle with sides’ length 3 and 4 has length 5.

2️⃣ Diagram by Sir Thomas Heath (20th century)

Puzzle difficulty: ☕️

Hint: This modern reconstruction illustrates a purely visual proof of the theorem itself.

🌱 Next Week

Next week, you’ll receive a freshly brewed selection of new puzzles — designed to both stimulate and delight your mathematical curiosity.

Until then, I wish you a beautiful and meaningful week!

With love and curiosity,
Sybille ☕️

✏️ …. Solutions ….

1️⃣ The Zhoubi Suanjing’s 3-4-5 Proof (c. 100 B.C.E.)

2️⃣ Rearrangement Proof of Pythagoras Theorem

Rearrangement proof attributed to Sir Thomas Heath (20th century).

Welcome back to the Math Café, where math meets its modern counterpart, AI.

This week, we explore how reasoning models are evolving from brilliant students into dependable tutors and assistants, as well as what that means for learning, research, and everyday curiosity… and most of all for you!

🎓 From “Gifted Student” to Reliable Personal Tutor

🧭 The last few weeks of 2025 marked a decisive moment in AI’s evolution.
Systems that once dazzled researchers with clever problem-solving are now maturing into disciplined, ready-to-use assistants.

At the University of Maine’s Responsible AI in K–12 Mathematics summit, educators are testing adaptive tutoring platforms like ALEKS for Calculus, which deliver personalized, context-aware learning.

🤖 AI can now serve as your own study companion — one that adapts to your level and goals.

Here’s one way to start your own “personal tutor” session:

💬 Prompt example
“You are a domain expert in [field]. I already know [related topics you’ve studied]. I want to [your goal]. Walk me through how to study efficiently.”

🧠 Reasoning Models Enter the Research Labs

AI is no longer just answering questions — it’s beginning to discover truths. 🌟

Recent scientific breakthroughs show what reasoning-capable systems can achieve:

• 🥇 Harmonic Team’s Aristotle combines informal reasoning with formal verification in Lean, achieving gold-medal-level performance on Olympiad math problems — and checking its own proofs.

• 🔢 O-Forge integrates large language models with computer algebra systems to to solve complex inequalities.

• 🇫🇷 Telecom SudParis and École Polytechnique researchers used GPT-5 to solve five of six IMO problems and even prove new theorems, producing Lean-verifiable proofs, this with minimal human intervention.

• 🌌 Oxford researchers, in collaboration with Gemini, classified stellar events — asteroid collisions, supernovae, and black hole disruptions — with 96.7% success explaining their reasoning along the way.

• ⚛️ At Los Alamos, the THOR system combines tensor networks with AI to solve solve PDEs in materials science.

  Together, these developments mark a shift:
  from using AI to assist thinking to building systems that can think alongside us. 🧩

⚙️ The Hardware Race Behind the Reasoning Revolution

💾 None of this would be possible without a tectonic shift in computing infrastructure.

AMD and OpenAI have launched a a multi-year, multi-billion-dollar chip partnership.

And Broadcom joined in, designing custom accelerators expected to exceed 10 GW of compute.

Together, they’re fueling an AI supercycle that’s driving semiconductor innovation, and making reasoning-capable AI models more accessible than ever. 🚀

🔧 Your Turn: How You Can Experiment Today

You don’t need a lab to explore reasoning AI; just curiosity and a few smart prompts. 💡

💭 To check a solution’s reliability

“You are an expert in [field]. Review the solution rigorously and point out any flaws.”

🎯 To learn a new skill

“You are a mentor guiding a new [role]. Teach me [task/workflow] step by step, flagging risks.”

Try these prompts, adapt them to your needs, and see how AI performs as a collaborator, not just a calculator. 🧮

So, what do you think?
Are you already working closely with AI, or just starting to explore it?

💬 Share your thoughts in the comments: I’d love to hear how you use (or challenge!) these tools.

🗓️ Next Week’s Teaser

Next Tuesday, the Café will dive deeper into one of the puzzles — with a twist you might not expect. 🕵️‍♀️

Until then, have a wonderful week.

With love and curiosity,
Sybille ☕

🧮 Detailed Solutions to Last Week’s Puzzles

Puzzle 1
❗ There is a division by zero!

Puzzle 2
📐 Bhāskara’s diagram offers a geometric proof of the Pythagorean Theorem.

Pick a right triangle, with hypothenuse of length c, and whose other sides are of length a (for the smallest one) and b (for the remaining one). Note that each triangle has area ab/2. Arrange four copies of this triangle in a square as shown in the initial diagram:

Note that the blue space in the middle of the figure is a square of area (b-a)². The four triangles and the blue square have total area equal to 2ab+(b-a)², which according to the diagram is the same as the area of the square of side c, i.e. is equal to c².

We obtain: 2ab+(b-a)²=c², which after simplification yields c²= a² + b², which is indeed the Pythagorean Theorem.📐

Three puzzles, three sips of math: Are you ready for your mid-October brain brew?

Welcome to this second edition of The Math Cafe!

I’m so glad to see you here.

A bit more about me: I’m Sybille, a mathematician specialised in Algebraic and Enumerative Geometry. I believe studying and doing math can be genuinely fun and stimulating when approached with curiosity; and my goal here is to share that spirit with you. I aim to make math friendlier and more approachable, and I’d love your feedback: what caught your attention, what felt too dry, or what could be clearer? You can contact me here, or by writing in the comments.

For today’s edition, I selected three puzzles from algebra, geometry and probability, each one trickier than the last. You’ll find the estimated difficulty marked with coffee cups ☕, a hint of how many cups you might need to tackle the problem (just a warning, not a recommendation!).

Note:

Puzzles 1 and 2 only require high-school algebra and geometry, and a touch of creativity.

Puzzle 3 calls for some probability basics, and sharper problem-solving instincts.

🧩 The Puzzles

Puzzle 1 — ☕

Can you spot the error in this ‘proof’?

A good warm-up for spotting logical slips: can you pinpoint the exact flaw?

Puzzle 2 — ☕☕

The diagram below was found in the writings of Indian mathematician Bhāskara II (12th century). It hides the proof of a famous theorem. Can you discover which one?

Hint: You probably learned the theorem in junior high school.

Note that Bhāskara’s elegant diagram offers a wordless proof: a perfect example of how geometry can speak for itself.

Puzzle 3 — ☕☕☕☕☕

Determine how many times, on average, one must toss a fair coin (independently and successively) until obtaining a sequence of an odd number of heads followed by a tail.

This problem appeared in the 2019 entrance exam for the elite mathematics program at ENS Ulm, Paris.

Candidates had only a blackboard, a piece of chalk, and 25 minutes to make decent progress on it.

So... would you get in? 🤓

🔹 Suggested approach: Try the puzzles in order of difficulty, and, most importantly, have fun!

That’s all for this week’s brew of puzzles! How did you find them: Too easy, too tricky, or just right? I’d love to hear from you!

🗓️ Next week teaser

Next Tuesday, The Café will serve a freshly brewed selection of recent news in the fields of Mathematics and AI.

I look forward to seeing you next Tuesday.

Until then, have a wonderful week!

With love and curiosity,

Sybille ☕

P.S.: Solutions will appear in next week’s edition of The Café.

In the meantime, paid subscribers are very welcome to share their solving process and tentative answers using this form: I’ll send back personal feedback, tips, and good coffee vibes ☕💡.

The Collatz Conjecture: When Simple Turns Impossible

Welcome to the very first edition of The Math Café!
I’m so glad to see you here.

On the menu today: a deceptively simple puzzle that has stumped some of the brightest mathematical minds of our time.

☕ Subscribe to The Math Café for weekly stories (modern research topics, news, and puzzles) about beautiful, mysterious mathematics.

Subscribe now

The Story

Eighty-eight years ago, a young mathematics assistant named Lothar Collatz noticed a curious pattern:

1. Take any positive number.

2. If it’s even - divide it by 2.

3. If it’s odd - triple it and add 1.

4. Repeat.

No matter where you start, you seem to always end up at 1.

For example:

• Start with 21 → 64 → 32 → 16 → 8 → 4 → 2 → 1.

• Start with 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.

Recommended: Try it with your birthday month. But beware - it might take longer than you expect (up to 19 steps)!

For example, if you were born in November (11): 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.

Some Good (and Bad) News

Looks simple enough, right?

The good news: If you manage to prove that this process always reaches 1, you could earn over $800,000 USD - just publish your proof in a recognized journal before 2031 and claim your prize from the Math Prize.

The bad news: No one has managed to prove it yet. No one can even be *sure* the conjecture is true.

Even Terence Tao, one of the greatest problem-solvers alive and a Fields Medalist, tried his hand at it. In 2019, he proved that the conjecture holds for almost all numbers (refining previous results by mathematicians Krasikov and Lagarias, Terras, and Korec). This extraordinary step forward involved sophisticated statistics tools, and great insights. But a complete proof, valid for all numbers, remains elusive.

On the other hand, massive computer searches have verified it for numbers up to sextillions (that’s billions of trillions!). Yet pure computation can never prove it true for every number, and only fail to find a counterexample.

Up to this date, numerous attempt involving AI-assisted formal proofs remain unsuccessful.

Why Is It So Hard?

The Collatz Conjecture lives in that tricky space where numbers behave chaotically.
Every simple step (divide by 2, or 3x+1) feels predictable - but the long-term outcome is almost impossible to foresee.

This makes it a perfect example of how complexity can emerge from simplicity, a recurring theme in mathematics.

What Became of Collatz?

Lothar Collatz went on to make important contributions to functional analysis, differential equations, and numerical analysis. Yet, to this day, he’s best remembered for this playful riddle from his early career - the one we now call the 3x+1 problem or the Collatz Conjecture.

So next time you face a puzzle that seems “too simple,” remember:
You might just be having your own Collatz moment. 🙂

☕ Your Turn

💭 What “deceptively simple” problem puzzled you this month?

Take a moment to reflect: sometimes, the hardest puzzles teach us the most 🪞.

Next week

Your first Espresso Shot: sharp puzzles to sharpen your logic.

Looking forward to seeing you there!

With curiosity and love, 🌿
Sybille for The Math Café

Pull up a chair, grab your favourite drink, and make yourself comfortable.
Breathe in the aroma of freshly brewed ideas — and settle in.

This little corner of the internet is the Math Café:
a place for the simply curious, the once-curious, or the secretly curious to gather around math.

But don’t worry — this isn’t a classroom.
No exams. No formulas. No stress.
Here, math is something to enjoy, not endure.

💡 Why a “Café”?

Because cafés are warm and friendly.
You don’t need an invitation — you just sit down and join the conversation.

That’s the atmosphere I want to create here:

☕ Short explorations of math curiosities — puzzles, quirky patterns, surprising stories.
🪶 Light peeks into research, to see how math is alive today.
💬 And above all: conversation. I’d love to hear what sparks your curiosity, too.

📅 What to Expect

Every Tuesday, you’ll get a free “cup of math” in your inbox — short, playful, and thought-provoking:

• A puzzle to tickle your logic.

• A glimpse of an open problem in modern research.

• A curious and beautiful proof — with pictures.

Think of it as espresso shots of math — small, energizing, and fun.

🔍 Coming Next Week

We’ll start with one of the simplest unsolved riddles in mathematics —
so simple you can try it at home,
so mysterious it’s resisted the world’s brightest minds for nearly a century.

💬 Join the Conversation

What’s your relationship with math?
Did you love it in school, avoid it, or fall back in love with it later?
Drop a note in the comments — I’d love to hear from you.

✨ Subscribe for Your Weekly Cup

New posts appear every Tuesday morning — perfect for your first sip of coffee.

👉 Subscribe to The Math Café to get them straight to your inbox.

Subscribe now

Paid subscribers also get:

• Behind-the-scenes notes and extra math stories.

• Early access to my ebook Most Beautiful Math Proofs.

• Bonus “refills” — extra posts and personal insights.

Welcome to the Math Café.
Curiosity is always brewing here. 🌿

With love,
Sybille