Cryptarithms

A cryptarithm is an arithmetic problem where digits have been disguised as letters. The classic example is: $\text{SEND} + \text{MORE} = \text{MONEY}.$

Each letter stands for one digit; the same letter is the same digit everywhere; different letters are different digits. The puzzle is to figure out what each letter is.

Idea

Cryptarithms are solved by patient logical reasoning, not random trial. Here are the common tools:

• First digit cannot be $0$. So in $\text{MONEY}$, $M \neq 0$.
• Carries are tiny. Adding two single digits gives at most $19$, so a column generates a carry of $0$ or $1$ (or $2$ when adding three).
• Use columns from the right. The units column is independent of all the others. Solve it first, then move left.
• Constraint of distinct digits. Once a digit is fixed, no other letter can use it.

For example, in the simpler puzzle $\text{ABC} + \text{ABC} = \text{BCC},$ the units column says $C + C$ ends in $C$, so $C = 0$. Then $B + B = C = 0$ (with carry), so $B + B = 10$, $B = 5$. Etc.

The thrill: when the last column is solved, every letter snaps into place.

Some cryptarithms allow multiple solutions; others have exactly one. A good puzzle always has just one.

Worked examples

Example 1. Solve $\text{A} + \text{A} = \text{BA}$, where letters are different digits.

Units: $A + A = A + 10 \cdot \text{carry}$. For the units of LHS to match units of RHS (which is $A$), we need $2A \mod 10 = A$, so $A = 0$ or carry happens. If $A = 0$, no carry, so $\text{BA} = 0$ , but $B$ leads a number so $B \neq 0$. Try carry: $2A = A + 10$, so $A = 10$ , impossible. So $A = 0$ , but then LHS $= 0$, RHS has $B$ leading, contradiction. Hence no solution. (A good check: the puzzle as stated cannot be satisfied.)

Example 2. $\text{AB} + \text{BA} = \text{CC}$. Find a solution.

LHS $= 10A + B + 10B + A = 11(A + B)$. RHS $= 11C$. So $A + B = C$. Many solutions: $A=1, B=2, C=3$ gives $12+21=33$ ✓. Or $A=2, B=5, C=7$ gives $25+52=77$ ✓.

Example 3. Solve $\text{ABC} + \text{ABC} + \text{ABC} = \text{BBB}$.

LHS $= 3 \times \text{ABC} = 333 \cdot B \div 1$. So $3(100A + 10B + C) = 111 B$. Simplify: $300A + 30B + 3C = 111B \Rightarrow 300A + 3C = 81B$. Divide by $3$: $100A + C = 27B$. With single digits: $A = 1, B = 4, C = 8$ gives $108 = 108$ ✓. So $\text{ABC} = 148$. Check: $148 \times 3 = 444 = \text{BBB}$ ✓.

Example 4. Solve $\text{ON} + \text{ON} + \text{ON} = \text{GO}$ (each letter a different digit).

LHS $= 3 \times \text{ON} = 3(10O + N) = 30O + 3N$. RHS $= 10G + O$. So $30O + 3N = 10G + O \Rightarrow 29O + 3N = 10G$. For $G$ single-digit, $29O$ must be small; try $O = 1$: $29 + 3N = 10G \Rightarrow 3N = 10G - 29$. $G=4$: $3N = 11$ no. $G=5$: $3N=21$, $N=7$. So $O=1, N=7, G=5$. Check: $17+17+17=51$ ✓.

Try it yourself

1. Solve $\text{AB} + \text{BA} = \text{CC}$ with $A=3, B=4$. What is $C$?
2. Solve $\text{ABC} - \text{CBA} = ?$. Take $A=8, B=2, C=1$.
3. Find all single-digit pairs $(A, B)$ with $A + B = 9$.
4. Solve $\text{X} \times 3 = \text{YX}$ (different digits).
5. Solve $\text{TWO} + \text{TWO} = \text{FOUR}$ (one solution exists; try by columns).
6. Decode: $\text{AA} \times \text{B} = \text{CDD}$ with $B = 7$.
7. Solve $\text{AB} + \text{BC} = \text{CA}$ for any small example.
8. Compute $\text{ABCABC} \div \text{ABC}$. What letter-puzzle is this?

Activity

Make your own cryptarithm. Take any addition fact, replace each digit with a letter (consistent across the equation), and challenge a friend to decode it.