🔢 Matthews Numbers — When Triangular Numbers Play the Splitting Game

Ages: 11–14  ·  Duration: 90–120 minutes  ·  Topics: Triangular Numbers, Digit Properties, Divisibility, Chinese Remainder Theorem  ·  Date: February 2026

Source: FSJM (Fédération Suisse des Jeux Mathématiques), Challenge 18 · Archives

Part 0 — Warm-Up: Stacking Bowling Pins (10 min)

The Big Idea

"You're setting up bowling pins. The first row has 1 pin, the second row 2, the third row 3… How many pins total when you've set up 4 rows?"

🗣️ Pair discussion: compute $1 + 2 + 3 + 4$.

The answer is $1 + 2 + 3 + 4 = 10$. These are called triangular numbers — they count the dots in triangular arrangements:

Row 1:    •           → T₁ = 1
Row 2:   • •          → T₂ = 1+2 = 3
Row 3:  • • •         → T₃ = 1+2+3 = 6
Row 4: • • • •        → T₄ = 1+2+3+4 = 10

Quick Practice (hands up — shout it out!)

✋ "Adding all those numbers gets tedious. Is there a shortcut?"

The Gauss Trick — A Shortcut Formula

There's a famous story about the mathematician Carl Friedrich Gauss. When he was about your age, his teacher asked the class to add up $1 + 2 + 3 + \cdots + 100$. While other students started adding one by one, Gauss found the answer almost instantly. Here's how:

✍️ Let's discover his trick together, using $n = 10$:

Write the sum forwards and backwards, one above the other:

Now add each column:

Every column sums to the same thing: $10 + 1 = 11$. And there are 10 columns.

So: $2S = 10 \times 11 = 110$, which means $S = \frac{110}{2} = 55$. ✅

✋ "Why does every column sum to 11?" Because each column pairs a number $k$ from the top with $11 - k$ from the bottom. Their sum is always $k + (11 - k) = 11 = n + 1$.

Key Rule

Why does this formula work? With the Gauss trick for any $n$:

• Write the sum twice (forwards and backwards).
• Each column sums to $n + 1$.
• There are $n$ columns.
• So $2 \times T_n = n \times (n+1)$, giving $T_n = \frac{n(n+1)}{2}$.

Let's verify with a few examples:

Part 1 — The Simplified Version: Digit Splitting (≈ 15–20 min)

Setting the Scene

Before today's main challenge, let's explore a curious idea: splitting a number's digits.

Rule: Take a number and cut its digit string into a left part and a right part. We say a number has a "double split" if the left part equals twice the right part.

Important constraint: The right part cannot start with 0. (Leading zeros don't make valid numbers!)

🧩 "Can you find some numbers with a double split?"

Worked Examples

Let's start with $21$. We split it as $2 \mid 1$. The left part is 2, the right part is 1. Is $2 = 2 \times 1$? Yes! ✅

Let's try $105$. We could split it as $1 \mid 05$, but 05 starts with a zero — not allowed! However, $10 \mid 5$ works: $10 = 2 \times 5$ ✅.

✍️ Work in pairs for 3 minutes: Find at least 3 more numbers with a double split.

The Key Insight — Structure of "Double-Split" Numbers

Let's figure out the algebra behind these numbers. We'll work through a concrete example first, then generalise.

Concrete example: 2211

The number $2211$ has 4 digits. We split it in the middle: left part $= 22$, right part $= 11$.

Now, how is $2211$ built from these two parts?

• The right part (11) occupies the last 2 digit positions.
• The left part (22) is shifted left by 2 places, which means multiplying by $10^2 = 100$.

So: $2211 = 22 \times 100 + 11$.

Since the left part is twice the right part ($22 = 2 \times 11$), let's call the right part $k$. Then:

Check: $k = 11$, and $11 \times 201 = 2{,}211$. ✅

General rule for any even digit count:

If a number $N$ has $2d$ digits total and we split it in the middle ($d$ digits on each side), with right part $= k$ and left part $= 2k$:

What range can $k$ take?

Both the left and right halves must have exactly $d$ digits. Let's think about what this means for $k$:

• The right part $k$ has exactly $d$ digits, so: $10^{d-1} \le k \le 10^d - 1$.
  For $d = 3$: $100 \le k \le 999$.
• The left part $2k$ must also have exactly $d$ digits, so: $10^{d-1} \le 2k \le 10^d - 1$.
  For $d = 3$: $100 \le 2k \le 999$, which gives $50 \le k \le 499$.

Combining both constraints for $d = 3$: $100 \le k \le 499$.

Why not $k = 500$? Because $2 \times 500 = 1000$, which is 4 digits, not 3. The total would be $1000|500 = 1{,}000{,}500$ — that's a 7-digit number, not 6!

🗣️ "But what if we split at a position that's not the midpoint?" Good question — we'll check this in Part 2.

Part 2 — The Real Challenge ⭐ (≈ 20–25 min)

The Problem

Matthews Numbers (FSJM, Challenge 18):

On his calculator, Mathias adds the natural whole numbers in order: $0 + 1 + 2 + 3 + 4 + \cdots$

Before typing each number (starting from the second), the calculator displays a provisional result:
$0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots$

Matthew notes all the provisional results that can be split into two numbers such that the first is twice the second — for example: 21, 105, 2211, 9045, …

Matthew then chooses a six-digit number from those he has noted.

What is this number? Find all the possibilities.

Note: The split cannot be made just before a digit 0.

Step-by-Step Guided Discovery

Step 1: What are the "provisional results"?

Look at what Mathias is doing on his calculator:

• Before typing 1: display shows $0$
• Before typing 2: display shows $0 + 1 = 1$
• Before typing 3: display shows $0 + 1 + 2 = 3$
• Before typing 4: display shows $0 + 1 + 2 + 3 = 6$
• Before typing 5: display shows $0 + 1 + 2 + 3 + 4 = 10$

These are the triangular numbers $T_n = \frac{n(n+1)}{2}$:

✋ "Let's verify: is $T_6 = 21$?"

Step 2: Verify the examples from the problem.

🧩 Challenge: check each one before I show the answer.

Step 3: Where can we split a 6-digit number?

A 6-digit number $N$ sits between 100,000 and 999,999. Its digits can be cut at 5 possible positions. Let's check each one carefully:

Split 1 | 5 (1 digit on the left, 5 on the right)

The left part $L$ has 1 digit, so $L$ is between 1 and 9. Since $L = 2R$, we get $R \le 4.5$. But $R$ has 5 digits, so $R \ge 10{,}000$. We can't have both! ❌ Impossible.

Split 2 | 4 (2 digits left, 4 right)

$L$ has 2 digits: $L \le 99$, so $R = L/2 \le 49$. But $R$ has 4 digits: $R \ge 1{,}000$. ❌ Impossible.

Split 3 | 3 (3 digits each side) ✅

$L$ has 3 digits: $100 \le L \le 999$. $R$ has 3 digits: $100 \le R \le 999$.

$L = 2R$ gives $100 \le 2R \le 999$, so $50 \le R \le 499$. Combined with $R \ge 100$: we need $100 \le R \le 499$. This works! ✅

Split 4 | 2 (4 digits left, 2 right)

$R$ has 2 digits: $R \le 99$, so $L = 2R \le 198$. But $L$ has 4 digits: $L \ge 1{,}000$. ❌ Impossible.

Split 5 | 1 (5 digits left, 1 right)

$R \le 9$, so $L = 2R \le 18$. But $L$ has 5 digits: $L \ge 10{,}000$. ❌ Impossible.

From Part 1 (with $d = 3$), this means:

and $N$ must also be a triangular number.

Dead-End: Why "Just Searching" Is Painful

One might try: "Let me check all multiples of 2001 from $2001 \times 100 = 200{,}100$ to $2001 \times 499 = 998{,}499$." That's 400 numbers to check by hand! We need smarter tools.

🗣️ "So what equation do we need to solve?"

We need $T_n = 2001k$, which gives $\frac{n(n+1)}{2} = 2001k$. Multiply both sides by 2:

So $n(n+1)$ must be divisible by $4002$. What are the prime factors of $4002$?

Factoring 4002 step by step:

• $4002$ is even, so $4002 = 2 \times 2001$.
• Is $2001$ divisible by $3$? The digit sum is $2 + 0 + 0 + 1 = 3$, and $3$ divides $3$. Yes! $2001 / 3 = 667$.
• Is $667$ divisible by small primes? Try $7$: $667 / 7 = 95.3$… No. Try $11$: $667 / 11 = 60.6$… No. Try $13$: no. Try $17$: no. Try $19$: no. Try $23$: $667 / 23 = 29$. Yes! ✅

So: $4002 = 2 \times 3 \times 23 \times 29$. Check: $23 \times 29 = 667$ ✅ and $3 \times 667 = 2001$ ✅.

Now the key insight: $n$ and $n+1$ are consecutive integers, so they share no common factor. (Why? If some number $d$ divides both $n$ and $n+1$, then $d$ divides $(n+1) - n = 1$. The only number that divides 1 is 1. So $d = 1$.)

Each prime factor $2, 3, 23, 29$ must go entirely to either $n$ or $n+1$ — it can't be "split" between them.

This is where the Chinese Remainder Theorem enters — but first, a break!

☕ Break (5 min)

Part 3 — Theory & Formalisation (≈ 15–20 min)

Tool 1: Testing Whether a Number Is Triangular

We've been checking "is $N$ triangular?" by trying to find an $n$ with $\frac{n(n+1)}{2} = N$. Is there a quicker test? Yes!

Let's derive it step by step.

Start: We want to solve $\frac{n(n+1)}{2} = N$ for $n$.

Step 1: Multiply both sides by 2:

Step 2: Expand the left side:

Step 3: Move everything to one side to get a standard quadratic equation $an^2 + bn + c = 0$:

Here $a = 1$, $b = 1$, $c = -2N$.

Step 4: Use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Let's simplify the expression under the square root (the discriminant):

So the formula becomes:

Step 5: Since $n$ must be a positive whole number, we take the $+$ sign (the $-$ sign gives a negative number):

For this to be a whole number, $1 + 8N$ must be a perfect square, and $\sqrt{1 + 8N}$ must be odd. (Actually the "odd" part comes for free: $8N$ is even, so $8N + 1$ is odd, and an odd perfect square always has an odd root. ✅)

Worked Example

Is $N = 222{,}111$ triangular?

• Compute $8 \times 222{,}111 + 1 = 1{,}776{,}888 + 1 = 1{,}776{,}889$.
• Is $1{,}776{,}889$ a perfect square? Estimate: $\sqrt{1{,}776{,}889} \approx 1{,}333$.
• Check: $1{,}333^2 = (1{,}300 + 33)^2 = 1{,}300^2 + 2 \times 1{,}300 \times 33 + 33^2$ $= 1{,}690{,}000 + 85{,}800 + 1{,}089 = 1{,}776{,}889$. ✅ It's a perfect square!
• So $n = \frac{-1 + 1{,}333}{2} = \frac{1{,}332}{2} = 666$.
• Verify: $T_{666} = \frac{666 \times 667}{2} = \frac{444{,}222}{2} = 222{,}111$. ✅

Tool 2: Why Consecutive Integers Are Coprime

Proof: Suppose some integer $d \ge 1$ divides both $n$ and $n+1$.

• Since $d$ divides $n$, we can write $n = d \times q_1$ for some integer $q_1$.
• Since $d$ divides $n + 1$, we can write $n + 1 = d \times q_2$ for some integer $q_2$.
• Subtract: $(n+1) - n = d \times q_2 - d \times q_1 = d \times (q_2 - q_1)$.
• The left side is $1$, so $d \times (\text{some integer}) = 1$.
• Since $d \ge 1$, the only possibility is $d = 1$. $\square$

Concrete examples:

• $n = 12$, $n+1 = 13$: factors of 12 are $\{1,2,3,4,6,12\}$; factors of 13 are $\{1,13\}$. Common: just $1$. ✅
• $n = 35$, $n+1 = 36$: factors of 35 are $\{1,5,7,35\}$; factors of 36 are $\{1,2,3,4,6,9,12,18,36\}$. Common: just $1$. ✅ Even though both are composite!

Why this matters: We need $4002 = 2 \times 3 \times 23 \times 29$ to divide $n(n+1)$. Since $\gcd(n, n+1) = 1$, no prime can divide both $n$ and $n+1$. Each prime factor goes entirely to one side.

Tool 3: What About the Factor of 2?

We need $\frac{n(n+1)}{2} = 2001k$. The denominator 2 might worry us, but actually $n(n+1)$ is always even — because of the two consecutive numbers $n$ and $n+1$, one is always even and the other odd.

Examples: $7 \times 8 = 56$ (8 is even). $10 \times 11 = 110$ (10 is even). So dividing by 2 always works. The factor of 2 takes care of itself.

What really matters: which primes from $2001 = 3 \times 23 \times 29$ divide $n(n+1)$?

Tool 4: The Chinese Remainder Theorem (CRT)

That's abstract. Let's see what it means concretely.

Worked CRT example: Find all $x$ such that $x \equiv 2 \pmod{3}$ and $x \equiv 1 \pmod{5}$.

This says: "$x$ leaves remainder 2 when divided by 3, and remainder 1 when divided by 5."

Method — list and match:

• Numbers with remainder 2 when divided by 3: $2, 5, 8, 11, 14, 17, 20, 23, 26, \ldots$
• Numbers with remainder 1 when divided by 5: $1, 6, 11, 16, 21, 26, 31, \ldots$
• First match: $\mathbf{11}$. Next: $11 + 15 = 26$, then $41$, etc.

So: $x \equiv 11 \pmod{15}$. The CRT guarantees one answer in each block of $3 \times 5 = 15$.

A second worked example (three moduli):

Find $x$ such that $x \equiv 0 \pmod{3}$, $x \equiv 22 \pmod{23}$, $x \equiv 0 \pmod{29}$.

Step 1: $3 \mid x$ and $29 \mid x$ means $87 \mid x$. So $x = 87, 174, 261, \ldots, 1218, \ldots$

Step 2: We need $x \equiv 22 \pmod{23}$, i.e., $x + 1 \equiv 0 \pmod{23}$.

• $87 + 1 = 88$. $88/23 = 3.83$. No.
• $174 + 1 = 175$. $175/23 = 7.6$. No.
• … (checking each $87m + 1$) …
• $\mathbf{1218 + 1 = 1219}$. $1219/23 = 53$. Yes! ✅

So $x \equiv 1218 \pmod{2001}$. (This exact value will appear as one of our solutions!)

How We'll Use CRT

For each prime $p$ in $\{3, 23, 29\}$, we need $p$ to divide either $n$ or $n+1$:

• "$p$ divides $n$" means $n \equiv 0 \pmod{p}$.
• "$p$ divides $n+1$" means $n+1 \equiv 0$, i.e., $n \equiv -1 \pmod{p}$:
  • For $p = 3$: $n \equiv 2 \pmod{3}$.
  • For $p = 23$: $n \equiv 22 \pmod{23}$.
  • For $p = 29$: $n \equiv 28 \pmod{29}$.

Each prime gives 2 choices. With 3 primes: $2^3 = 8$ systems. CRT gives a unique $n \pmod{2001}$ for each.

Solution 1 — Brute-Force Enumeration Bronze

Setting Up the Range

We need $100{,}000 \le T_n \le 999{,}999$.

Lower bound: $\frac{n(n+1)}{2} \ge 100{,}000 \implies n(n+1) \ge 200{,}000$. Estimate: $n \approx \sqrt{200{,}000} \approx 447.2$.

• $T_{446} = \frac{446 \times 447}{2} = 99{,}681$ — only 5 digits. ❌
• $T_{447} = \frac{447 \times 448}{2} = 100{,}128$ — 6 digits! ✅

Upper bound: $\frac{n(n+1)}{2} \le 999{,}999 \implies n(n+1) \le 1{,}999{,}998$. Estimate: $n \approx 1{,}414.2$.

• $T_{1413} = \frac{1413 \times 1414}{2} = 998{,}991$ — 6 digits. ✅
• $T_{1414} = \frac{1414 \times 1415}{2} = 1{,}000{,}405$ — 7 digits! ❌

The Search

Check all $N = 2001k$ for $k \in [100, 499]$: is $8N + 1$ a perfect square?

All other $k$ values in $[100, 499]$ fail.

Verification

Answer 1: $T_{666}$

$T_{666} = \frac{666 \times 667}{2} = \frac{444{,}222}{2} = 222{,}111$. Split: $222 \mid 111$. Is $222 = 2 \times 111$? Yes! ✅

Answer 2: $T_{782}$

$T_{782} = \frac{782 \times 783}{2} = \frac{612{,}306}{2} = 306{,}153$. Split: $306 \mid 153$. Is $306 = 2 \times 153$? Yes! ✅

Answer 3: $T_{1218}$

$T_{1218} = \frac{1218 \times 1219}{2} = \frac{1{,}484{,}742}{2} = 742{,}371$. Split: $742 \mid 371$. Is $742 = 2 \times 371$? Yes! ✅

Answer 4: $T_{1334}$

$T_{1334} = \frac{1334 \times 1335}{2} = \frac{1{,}780{,}890}{2} = 890{,}445$. Split: $890 \mid 445$. Is $890 = 2 \times 445$? Yes! ✅

Solution 2 — Algebraic Modular Analysis Silver

Key Observation

We need $2001 = 3 \times 23 \times 29$ to divide $T_n$. Since $\gcd(n, n+1) = 1$, each prime divides either $n$ or $n+1$.

Setting Up the 8 Systems

Solving the Systems (detailed)

System 1: All "0" — $n \equiv 0$ (mod 3, 23, 29). So $2001 \mid n$, giving $n = 0$ or $2001$. Both outside $[447, 1413]$. ❌

System 2: $n \equiv 0$ (mod 3), $n \equiv 0$ (mod 23), $n \equiv 28$ (mod 29)

$69 \mid n$. Need $n \equiv 28 \pmod{29}$. Solving: $n = 69 \times 21 = 1{,}449$. Outside range. ❌

System 3: $n \equiv 0$ (mod 3), $n \equiv 22$ (mod 23), $n \equiv 0$ (mod 29)

$87 \mid n$ (since $3 \mid n$ and $29 \mid n$). Need $n + 1$ divisible by 23.

Check multiples of 87: $522, 609, 696, 783, 870, 957, 1044, 1131, \mathbf{1218}, 1305, 1392$.

$1218 + 1 = 1219$. $1219 / 23 = 53$. ✅ → $n = 1{,}218$.

$T_{1218} = 742{,}371$. $k = 371$. Split: $742 \mid 371$. ✅

System 4: $n \equiv 0$ (mod 3), $n \equiv 22$ (mod 23), $n \equiv 28$ (mod 29)

$23 \mid (n+1)$ and $29 \mid (n+1)$: so $667 \mid (n+1)$. $n+1 = 667 \implies n = 666$, $n+1 = 1334 \implies n = 1333$.

Check $3 \mid n$: $666/3 = 222$ ✅; $1333$: digit sum $= 10$, not divisible by 3. ❌

$n = 666$. $T_{666} = 222{,}111$. $k = 111$. Split: $222 \mid 111$. ✅

System 5: $n \equiv 2$ (mod 3), $n \equiv 0$ (mod 23), $n \equiv 0$ (mod 29)

$667 \mid n$. In range: $n = 667$ or $1334$.

Check $n \equiv 2 \pmod{3}$: $667 = 3 \times 222 + 1 \equiv 1$. No. $1334 = 3 \times 444 + 2 \equiv 2$. ✅

$n = 1{,}334$. $T_{1334} = 890{,}445$. $k = 445$. Split: $890 \mid 445$. ✅

System 6: $n \equiv 2$ (mod 3), $n \equiv 0$ (mod 23), $n \equiv 28$ (mod 29)

$87 \mid (n+1)$ [since $3 \mid (n+1)$ and $29 \mid (n+1)$]. Also $23 \mid n$.

Check $n = 87t - 1$ for $23 \mid n$: $t = 9$ gives $n = 782$. $782/23 = 34$. ✅

$n = 782$. $T_{782} = 306{,}153$. $k = 153$. Split: $306 \mid 153$. ✅

System 7: $n \equiv 2$ (mod 3), $n \equiv 22$ (mod 23), $n \equiv 0$ (mod 29)

$29 \mid n$, $69 \mid (n+1)$. $n = 551$: $552/69 = 8$. ✅ But…

$T_{551} = 152{,}076$. $k = 76 < 100$. Split: $152 \mid 076$. Leading zero! ❌

System 8: All "-1" — $2001 \mid (n+1)$. $n = 2000$. $T_{2000} = 2{,}001{,}000$ — 7 digits. ❌

Summary Table

Solution 3 — Factor-Assignment Method Gold

Setup

Instead of CRT, we directly assign prime factors of $4002 = 2 \times 3 \times 23 \times 29$ to $n$ or $n+1$.

Let $A$ = product of primes assigned to $n$, $B$ = product assigned to $n+1$. Then $A \mid n$, $B \mid (n+1)$, and $A \times B = 4002$.

Case-by-Case Search

Case A: $6 \mid n$ and $667 \mid (n+1)$.

$n + 1 = 667 \implies n = 666$. Is $6 \mid 666$? $666/6 = 111$. ✅

$T_{666} = 222{,}111$. $k = 111$. Split: $222 \mid 111$. ✅

Case B: $46 \mid n$ and $87 \mid (n+1)$.

Check multiples of 87 in range: $783 \implies n = 782$. $782/46 = 17$. ✅

$T_{782} = 306{,}153$. $k = 153$. Split: $306 \mid 153$. ✅

Case C: $174 \mid n$ (factors $\{2,3,29\}$) and $23 \mid (n+1)$.

Multiples of 174 in range: $522, 696, 870, 1044, \mathbf{1218}$.

$1218 + 1 = 1219$. $1219/23 = 53$. ✅

$T_{1218} = 742{,}371$. $k = 371$. Split: $742 \mid 371$. ✅

Case D: $1334 \mid n$ and $3 \mid (n+1)$.

Only $n = 1334$ in range. $1335/3 = 445$. ✅

$T_{1334} = 890{,}445$. $k = 445$. Split: $890 \mid 445$. ✅

Case E: $29 \mid n$ and $138 \mid (n+1)$.

$n = 551$: $552/138 = 4$. ✅ But $T_{551} = 152{,}076$, $k = 76 < 100$. Leading zero! ❌

All remaining assignments give $n$ outside $[447, 1413]$.

Solution 4 — The Perfect-Square Reformulation Diamond (optional)

Setup

Let $m = 2n + 1$. Then $n(n+1) = \frac{m^2 - 1}{4}$ and $T_n = \frac{m^2 - 1}{8}$.

Quick check: $n = 6 \implies m = 13$. $\frac{169 - 1}{8} = \frac{168}{8} = 21 = T_6$. ✅

Derivation: $n^2 + n = n(n+1)$. We "complete the square":

Multiply by 4: $4n^2 + 4n = (2n+1)^2 - 1 = m^2 - 1$.

So $T_n = \frac{n(n+1)}{2} = \frac{m^2-1}{8}$.

The Equation

We need $\frac{m^2 - 1}{8} = 2001k$, giving $m^2 = 16{,}008k + 1$, i.e., $m^2 \equiv 1 \pmod{16{,}008}$.

$16{,}008 = 2^3 \times 3 \times 23 \times 29$. By CRT, this splits into conditions modulo 8, 3, 23, 29:

• $m^2 \equiv 1 \pmod{8}$: $m$ is odd (always true since $m = 2n+1$). ✅
• $m^2 \equiv 1 \pmod{3}$: $m \equiv 1$ or $2 \pmod{3}$. (2 choices)
• $m^2 \equiv 1 \pmod{23}$: $m \equiv 1$ or $22 \pmod{23}$. (2 choices)
• $m^2 \equiv 1 \pmod{29}$: $m \equiv 1$ or $28 \pmod{29}$. (2 choices)

This gives $1 \times 2 \times 2 \times 2 = 8$ residue classes for $m$ modulo 16,008.

Range: $100 \le k \le 499$ gives $m \in [1{,}266,\; 2{,}826]$ approximately.

Summary — The Answer

The Four Six-Digit Matthews Numbers

Why Are There Exactly 4 (and Not 8)?

Of the 8 possible CRT solutions modulo 2001, four are eliminated:

1. $n = 0$ and $n = 2001$: $T_n$ is not a 6-digit number (too small / out of range).
2. $n = 1449$ and $n = 2000$: $T_n$ has 7 digits (too large).
3. $n = 551$: the split gives $152 \mid 076$ — leading zero disqualifies it.

Complete List of Matthews Numbers (all digit lengths)

Why Multiple Solutions?

Extensions & Challenge Problems 🧩

🥉 Bronze

1. Five-digit Matthews numbers. Are there any 5-digit triangular numbers with a valid double split? Hint: Check all 4 possible splits (1|4, 2|3, 3|2, 4|1) using the size argument from Part 2.
2. Three-digit check. Verify that $T_{14} = 105$ is the only 3-digit Matthews number. Hint: for a 1|2 split, the structure constant is $21$.

🥈 Silver

3. Seven digits. Verify that $T_{1449} = 1{,}050{,}525$ is a Matthews number. Find all 7-digit Matthews numbers.
4. Generalise the constant. Factorise $20{,}001$ and $200{,}001$. What does this tell you about 8- and 10-digit Matthews numbers?

🥇 Gold

5. Odd-digit impossibility? Show algebraically that no split position allows a valid 5-digit double split.
6. The density question. As $d \to \infty$, does the count of $2d$-digit Matthews numbers grow, stay constant, or vanish?

🎓 Diamond (optional)

7. Triple split. Define $A|B|C$ with $A = 3C$, $B = 2C$. Find all triangular triple-split numbers up to 9 digits.
8. General ratio. Replace "twice" with "$r$ times" ($r \ge 2$). For which $r$ do infinitely many triangular $r$-split numbers exist?

Answer Key (for instructors)

This lecture was generated for a Math Circle session. See .copilot-instructions.md for authoring rules.

Source: FSJM, 38th competition, Challenge 18. Archives