Cryptarithms

$SEND + MORE = MONEY$ looks like nonsense until you remember it is not really about letters at all — it is arithmetic wearing a disguise. A cryptarithm replaces the digits of a correct sum with letters, and the puzzle is to unmask each letter as a digit from 0 to 9 so the sum becomes true again. Every letter keeps one fixed digit throughout, different letters get different digits, and no number is allowed to start with a hidden zero. Class 8's Number Play chapter treats cryptarithms as the payoff round for everything else in the chapter — place value, parity, and divisibility reasoning all get reused here, just aimed at a new kind of question: not "prove a property" but "find the hidden digits." The tool that cracks every cryptarithm is the same one behind long-division and column-addition checks — writing numbers as $100h+10t+u$ and reading off what each column's arithmetic forces.

The core insight

A cryptarithm is solved by rewriting each letter-pattern as a place-value equation and analysing it column by column from the units digit outward, using the facts that a carry between columns is always 0 or 1 and that no leading digit is 0.

The core idea

Every cryptarithm rests on four rules. Each letter is exactly one digit; the same letter always means the same digit; different letters mean different digits (unless stated otherwise); and a number's leading digit is never 0, since $0AB$ would just be the shorter number $AB$ .

Translate the picture into algebra. Any multi-digit number written in letters is really a place-value sum. A three-digit number $\overline{PQR}$ stands for

\overline{PQR} = 100P + 10Q + R,

where: $P$ , $Q$ , $R$ are the digits in the hundreds, tens and units places. This single idea turns a picture-puzzle into an equation you can actually solve — instead of guessing letters, you manipulate numbers.

Work from the units column outward. The rightmost column is the only one whose sum does not depend on a carry from somewhere else, so it is always the safest place to start. Once that column's digits and carry are pinned down, they feed into the next column, and so on leftward — exactly the order you already add numbers on paper, just run in reverse to discover the digits rather than combine known ones.

Bound the carries. Adding two single digits gives at most $9+9=18$ , so any carry passed into the next column can only be $0$ or $1$ — never more. This small fact is what makes a seemingly endless guessing game into a short, systematic search.

Figure

A column addition of two three digit numbers CAT plus CAT showing carry arrows moving from the units column to the tens column to the hundreds column

Derivation ATurning a letter pattern into a solvable place-value equation

The goal is to show, with nothing beyond the place-value expansion and basic algebra, how a picture-puzzle like $AA + AA = BAC$ becomes a single equation in whole-number unknowns that can be pinned down using the size limits on digits — this is the reasoning engine behind every cryptarithm solution, not just this one example.

Step 1. Expand every letter-block using place value. The two-digit number $AA$ has the same digit $A$ in both places, so $AA = 10A + A = 11A$ . The three-digit number $BAC$ has hundreds digit $B$ , tens digit $A$ , and units digit $C$ , so $BAC = 100B + 10A + C$ .

Step 2. Write the puzzle as an equation of these expansions. The sum $AA + AA$ becomes $11A + 11A = 22A$ , and it must equal $BAC$ , so $22A = 100B + 10A + C$ .

Step 3. Simplify to isolate the unknowns cleanly. Subtracting $10A$ from both sides gives $12A = 100B + C$ . This is now a pure number equation: the left side is at most $12 \times 9 = 108$ (since $A$ is a single digit), which immediately tells us $B$ can only be $0$ or $1$ — and $B=0$ is forbidden because $BAC$ cannot have a leading zero, so $B=1$ is forced before any guessing begins.

Step 4. Use the bound to finish. With $B=1$ , the equation becomes $12A = 100 + C$ , and since $C$ is a single digit ( $0$ to $9$ ), $12A$ must lie between $100$ and $109$ . Testing digits gives $A=9$ : $12(9)=108=100+8$ , so $C=8$ . Checking distinctness, $A=9,B=1,C=8$ are all different, and the leading digit $B=1$ is nonzero, so the algebra has produced a fully valid answer without trial and error over all ten digits. Verified: $AA=99$ , so $99+99=198$ , and reading $BAC$ as $B=1,A=9,C=8$ gives exactly $198$ .

Worked example

Solve the puzzle $CAT + CAT = DOG$ , given the extra clue $C=1$ and that all six letters stand for distinct digits.

Step 1. Set up the place-value equation. Since $C=1$ , $CAT = 100 + 10A + T$ , and doubling gives $DOG = 200 + 20A + 2T$ . Because $DOG$ is a three-digit number, this stays under $1000$ , which is automatically satisfied since $C=1$ keeps $CAT$ well under $500$ .

Step 2. Solve the units column first. Doubling $T$ must end in digit $G$ , possibly carrying $1$ into the tens column: $2T = 10\times(\text{carry}) + G$ .

Step 3. Solve the tens column next, folding in the units carry: $2A + \text{carry} = 10\times(\text{new carry}) + O$ .

Step 4. Solve the hundreds column last: $2C + \text{carry from tens} = D$ , and since $C=1$ , this is $D = 2$ (no carry) or $D=3$ (carry of $1$ ).

Step 5. Try values consistent with all three columns. Take $A=3$ : without an incoming carry, tens gives $2(3)=6$ , so $O=6$ , and no carry passes to hundreds, giving $D=2$ . Take $T=4$ : units gives $2(4)=8$ , so $G=8$ , with no carry into tens — consistent with the tens step above.

Step 6. Verify every digit is distinct and the sum checks out. The digits are $C=1, A=3, T=4, G=8, O=6, D=2$ — all six different. Substituting back, $CAT = 134$ , and $134+134=268$ , which reads as $D=2, O=6, G=8$ — exactly $DOG$ .

The solution is $C=1, A=3, T=4, D=2, O=6, G=8$ , giving $134+134=268$ .

Competitive stretch

If instead the clue were " $G$ is even and greater than $6$ ," the units-column equation $2T = 10(\text{carry})+G$ still applies, but now $G \in \{8\}$ forces $2T$ to end in $8$ , so $T=4$ or $T=9$ — the same column logic, just used to narrow choices before hundreds are touched, rather than after guessing $A$ and $T$ freely.

Common traps

Letting a letter shift value partway through the puzzle — the same letter must mean the same digit in every appearance, in every number.
Allowing a result's leading digit to be 0, silently turning a three-digit answer into a two-digit one and breaking the puzzle's stated form.
Dropping a carry when moving from one column to the next — a carry of 1 from units must be added into the tens-column equation, not ignored.
Guessing digits at random instead of using the place-value equation and the "carry is 0 or 1" bound to eliminate most possibilities before testing any.

Ask the tutor: Want to try cracking the classic $\text{SEND}+\text{MORE}=\text{MONEY}$ puzzle together, step by step?

How do I know where to start when a cryptarithm looks completely tangled?

Always start at the rightmost, units column — it is the only column whose equation does not depend on a carry from somewhere else, so it gives you a foothold with no unknowns hiding upstream. Once you fix the units digits and note whether they produce a carry, move one column left and repeat, carrying that carry forward. Working left to right, or jumping around, usually means juggling two unknown carries at once, which is far harder than necessary.

What if a puzzle seems to have more than one valid digit assignment?

Check both against every rule, not just the arithmetic: the same letter must be the same digit everywhere, different letters must be different digits, and no leading digit can be zero. Many "second solutions" fail one of these rules on closer inspection. If two assignments truly satisfy all four rules and the given sum, the puzzle as stated is ambiguous — well-posed cryptarithms are built so exactly one assignment survives, so double-check the original letter pattern for a missed clue.

Is there a faster way than testing digits one by one?

Yes — before testing anything, use the size bound the place-value equation gives you. In the $AA+AA=BAC$ case, the equation $12A=100B+C$ instantly rules out every value of $B$ except $1$ , since $12A$ can be at most $108$ . That single deduction shrinks the search from ten possible digits for $A$ down to checking just a couple that make $12A$ land between $100$ and $109$ . Always squeeze the equation for such bounds before trying digits.

Cryptarithms

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads