🏆 Bravo to Future Winners — A Cryptarithm Challenge

Math Circle Lecture  |  Ages 11–14  |  Duration: 90 minutes  |  Topics: Cryptarithms, place-value algebra, carry analysis, modular arithmetic
Source: FSJM 38th competition, Challenge 13  |  Date: 2026-02-08

Part 0 — Warm-Up: Secret Codes in Arithmetic (10 min)

The Big Idea

"What if every digit in a maths problem were replaced by a letter — and your job was to crack the code?"

🗣️ "Has anyone ever seen a coded addition before? Maybe in a puzzle book or a spy movie?"

A cryptarithm (or alphametic) is a puzzle where digits are replaced by letters:

• Each letter stands for exactly one digit (0–9).
• Different letters stand for different digits.
• No number starts with the digit 0.

The most famous cryptarithm was published in 1924:

  S E N D
+ M O R E
---------
M O N E Y

We won't solve that one today — we have a different challenge. But first, let's make sure we understand how these puzzles work.

Quick Practice (hands up — shout it out!) ✋

✍️ "Work with your neighbour for 2 minutes. Which value of $C$ makes $AB + BA = CCC$?"

Key Rule

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

Setting the Scene

Before tackling a 5-digit puzzle, let's try a smaller one.

Mini-Puzzle: Find distinct digits $P, Q, R, S$ (no leading zeros) such that $$PQ + RQ = SSS$$ and maximise $PQ$.

Wait — $PQ$ is a 2-digit number and $RQ$ is a 2-digit number, but $SSS$ is a 3-digit number. Can two 2-digit numbers even sum to a 3-digit number?

The maximum of two 2-digit numbers is $99 + 99 = 198$. And $SSS$ ranges from $111$ to $999$. So $SSS$ can only be $111$ (since $198 < 222$).

Therefore $S = 1$ and $PQ + RQ = 111$.

Worked Example

Write the column equations:

From the hundreds column: $c_2 = 1$, so $P + R + c_1 = 11$.

From the ones column: $2Q = 1 + 10c_1$.

• $c_1 = 0$: $Q = 0.5$ — not an integer!
• $c_1 = 1$: $Q = 5.5$ — not an integer!

🧩 "Wait — neither works? What happened?"

The Key Insight

This is a feature, not a bug. It shows us that the parity (even/odd) of the repeated letter is a powerful constraint.

Now let's try a solvable version!

Mini-Puzzle 2: Find distinct digits $A, B, C$ (no leading zeros) with $$AA + BB = CCC$$

Column equations:

• Ones: $A + B = C + 10c_1$
• Tens: $A + B + c_1 = C + 10c_2$
• Hundreds: $c_2 = C$

From hundreds: $c_2 = C$. Since $c_2 \in \{0, 1\}$, we need $C = 0$ or $C = 1$. $C = 0$ gives $CCC = 000$ (not valid). So $C = 1$ and $c_2 = 1$.

From tens: $A + B + c_1 = 11$. From ones: $A + B = 1 + 10c_1$.

• $c_1 = 0$: $A + B = 1$. Then tens: $A + B = 11$. Contradiction!
• $c_1 = 1$: $A + B = 11$. Then tens: $A + B + 1 = 11$ ✓.

So $A + B = 11$ with $A, B$ distinct, non-zero, and $\ne 1$. Pairs: $(2, 9), (3, 8), (4, 7), (5, 6)$.

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

The Problem

FSJM Challenge 13 — "Bravo to Future Winners" $$\begin{array}{r} G\;R\;A\;N\;D \\ +\; B\;R\;A\;V\;O \\ \hline X\;X\;X\;X\;X \end{array}$$ Each letter represents a unique digit (0–9). No number begins with 0.

What is the largest possible value of GRAND?

🗣️ "Take 30 seconds to count the letters. How many distinct digits do we need?"

The nine distinct letters are: $G, R, A, N, D, B, V, O, X$. Since there are only 10 digits (0–9), exactly one digit is unused.

Step-by-Step Guided Discovery

Step 1: What is $XXXXX$?

✋ "If all five digits of the sum are the same letter $X$, what kind of number is $XXXXX$?"

So the sum is a repdigit multiple of $11{,}111$, and $X \in \{1, 2, \ldots, 9\}$.

Step 2: Column Equations

Label the carries $c_1, c_2, c_3, c_4$ from right to left (each is 0 or 1):

Notice something special: the hundreds and thousands columns both involve a letter added to itself (because $A$ and $R$ appear in the same position in both GRAND and BRAVO).

Step 3: Parity Locks Everything Down

🧩 "Challenge: look at the hundreds column $2A + c_2 = X + 10c_3$ and the thousands column $2R + c_3 = X + 10c_4$. What does parity tell us?"

From hundreds: $X = 2A + c_2 - 10c_3$. Since $2A$ and $10c_3$ are even, $X$ has the same parity as $c_2$.

From thousands: $X = 2R + c_3 - 10c_4$. Similarly, $X$ has the same parity as $c_3$.

✍️ "Work in pairs: try both cases and see what values of $A$, $R$, and $X$ emerge."

Dead-End: Why Odd $X$ Fails

Case $c_2 = c_3 = 1$ (X odd):

• Hundreds: $2A + 1 = X + 10$, so $A = \frac{X + 9}{2}$.
• Thousands: $2R + 1 = X + 10c_4$.
  • If $c_4 = 0$: $R = \frac{X - 1}{2}$.
  • If $c_4 = 1$: $R = \frac{X + 9}{2} = A$ — clash!

So $c_4 = 0$ is forced, giving $R = \frac{X-1}{2}$ and $A = R + 5$.

Then the ten-thousands column: $G + B = X$.

The tens column gives $N + V + c_1 = X + 10$, so $N + V = X + 10 - c_1$.

Conclusion: $X$ must be even. ✓

☕ Break (5 min)

Part 3 — Theory & Formalisation (≈ 10–15 min)

The Technique: Column-Equation Analysis with Carries

Key deductions:

1. Parity propagation. If the same letter appears in the same column of both addends, the column equation becomes $2L + c_{\text{in}} = X + 10c_{\text{out}}$, which forces $X \equiv c_{\text{in}} \pmod{2}$.
2. Carry bounding. Since each digit $\le 9$, the maximum column sum is $9 + 9 + 1 = 19$, so carries are always 0 or 1.
3. Leading-digit constraint. The leftmost column has no outgoing carry (otherwise the sum would have more digits), so $G + B + c_4 = X$ exactly.

Even-$X$ Analysis (our puzzle)

With $c_2 = c_3 = 0$:

• $c_4 = 0 \Rightarrow R = A$ — clash!
• $c_4 = 1 \Rightarrow R = \frac{X}{2} + 5$, and $A = \frac{X}{2}$.

Mini-Exercise (pair work) ✍️

1. For $X = 8$, $A = 4$, $R = 9$, $G = 6$, $B = 1$: list the available digits for $N, D, V, O$ and show that $D + O = 8$ and $N + V = 8$ cannot both be satisfied.
2. For $X = 6$, $A = 3$, $R = 8$, $G = 4$, $B = 1$: find all valid $(N, D, V, O)$ from $\{0, 2, 5, 7, 9\}$.

Solution 1 — Carry-Chain Deduction 🥉 Bronze

Prerequisites: place-value notation, addition with carries.

Setup

We work column by column, right to left, tracking each carry.

Computation

Fact: $XXXXX = 11{,}111 \times X$ where $X \in \{1, \ldots, 9\}$.

Parity argument (from Part 3): $X$ must be even, $c_2 = c_3 = 0$, $c_4 = 1$.

This gives $A = X/2$, $R = X/2 + 5$, $G + B = X - 1$.

Checking each even $X$:

Only $X = 6$ survives. With $(G,B) = (4,1)$ for maximum $G$:

The available digits $\{0, 2, 5, 7, 9\}$ must fill $N, D, V, O$ with:

• $D + O = 16 \Rightarrow \{D, O\} = \{7, 9\}$
• $N + V = 5 \Rightarrow \{N, V\} = \{0, 5\}$

To maximise $GRAND = 48{,}3ND$, choose $N = 5$ (not 0) and $D = 9$ (not 7):

Result

Verification

Digit check:

Nine distinct digits, unused digit is 2. No leading zeros ($G = 4$, $B = 1$). ✓

Solution 2 — Algebraic Shortcut 🥈 Silver

Prerequisites: basic algebra, divisibility.

Key Observation

Since $R$ and $A$ appear in matching positions of both addends:

Derivation

Let $s = G + B$, $t = N + V$, $u = D + O$. Substituting $A = X/2$ and $R = X/2 + 5$:

Then:

With $s = X - 1$ (from the leading column with carry):

For $X = 6$: $10t + u = 66$.

Need $u \equiv 6 \pmod{10}$, so $u = 6$ or $u = 16$.

• $u = 6$: no pair from $\{0, 2, 5, 7, 9\}$ sums to 6.
• $u = 16$: $\{7, 9\}$ ✓. Then $t = 5$: $\{0, 5\}$ ✓.

Result

Solution 3 — Modular Arithmetic Approach 🥇 Gold

Prerequisites: modular arithmetic (mod 9, mod 11).

Setup

We use two classical divisibility tests:

Derivation

Mod 9 analysis:

The digit sum of $XXXXX$ is $5X$. The digit sum of $GRAND$ is $G + R + A + N + D$, and of $BRAVO$ is $B + R + A + V + O$.

If the unused digit is $m$, the total of all 9 assigned digits is $45 - m$. Since $R$ and $A$ are counted twice:

With $X = 6$, $A = 3$, $R = 8$: $56 - m \equiv 30 \equiv 3 \pmod{9}$.

Verification: $m = 2$, and $56 - 2 = 54 \equiv 0 \pmod{9}$. Hmm — let's recompute more carefully. Digit sum of GRAND: $4 + 8 + 3 + 5 + 9 = 29$. Digit sum of BRAVO: $1 + 8 + 3 + 0 + 7 = 19$. Sum: $48 \equiv 3 \pmod{9}$. Digit sum of $66{,}666$: $30 \equiv 3 \pmod 9$. ✓

Mod 11 analysis:

$11{,}111 = 11 \times 1{,}010 + 1$, so $XXXXX \equiv X \pmod{11}$.

Alternating sums: $(G + B) - 2R + 2A + (D + O) - (N + V) \equiv X \pmod{11}$.

Substituting: $5 - 16 + 6 + 16 - 5 = 6 \equiv 6 \pmod{11}$. ✓

Equivalently: $(D + O) - (N + V) = 16 - 5 = 11 \equiv 0 \pmod{11}$, which is consistent.

Result

Summary — The Answer

All Solutions

All 8 solutions share $X = 6$, $A = 3$, $R = 8$, unused digit $= 2$.

Why Multiple Solutions?

Extensions & Challenge Problems 🧩

🥉 Bronze

1. Find all solutions to the cryptarithm $AB + BA = CDC$ where $A, B, C, D$ are distinct digits. (Hint: $AB + BA = 11(A+B)$.)
2. In the GRAND + BRAVO puzzle, what is the smallest possible value of GRAND?

🥈 Silver

3. Prove that in any cryptarithm $PQRST + PQRST = XXXXX$, the value $X$ must be even and $PQRST$ determines $X$ uniquely.
4. How many essentially distinct solutions does the puzzle have if we only count assignments where $GRAND > BRAVO$?

🥇 Gold

5. Replace $XXXXX$ with $YYYYY$ and allow the sum to be a 6-digit repdigit: $GRAND + BRAVO = YYYYYY$. Is this possible? If so, find all solutions.
6. Create your own 5-letter + 5-letter = 5-digit cryptarithm where the two words share no letters (unlike GRAND/BRAVO which share $R$ and $A$). How does the difficulty change?

🎓 Diamond (optional)

7. Write a general solver: given any cryptarithm $W_1 + W_2 = W_3$ (arbitrary-length words), enumerate all solutions. What is the time complexity in terms of the number of distinct letters?
8. Prove that for any repdigit target $\underbrace{XX\cdots X}_{n}$, the number of cryptarithm solutions with two $n$-digit addends sharing $k$ letter positions is maximised when $k \approx n/2$.

Answer Key (for instructors)