What Is Cryptarithmetic?
Cryptarithmetic (also known as alphametics or verbal arithmetic) is a mathematical puzzle where standard decimal digits ($0$ to $9$) are replaced with letters or symbols in an arithmetic equation. The solver must assign a unique numerical digit to each distinct letter such that the entire equation remains mathematically sound, column by column.
Deduction, Not Random Guessing
The fundamental mental model of cryptarithmetic is constraint-driven deduction. Random trial and error on 10 letters yields $10! = 3,628,800$ permutations. By contrast, mathematical boundaries—such as leading carries ($M = 1$), column overflow ($k \le 1$), parity (even/odd), and units cyclicity—immediately eliminate over $99.9\%$ of possibilities, isolating the solution in a handful of logical steps.
Cryptarithmetic tests pure deductive reasoning, number sense, and simultaneous equation handling under time pressure. It separates brute-force guessers from structured logical thinkers.
A recurring feature in CAT LRDI puzzle sets, XAT decision reasoning, MAH MBA CET analytical puzzles, SNAP, NMAT, and premier campus recruitments (Infosys & TCS analytical rounds).
Decimal place-value expansion ($ABCD = 1000A + 100B + 10C + D$), basic single-digit column arithmetic, carry bounds ($0 \le k \le 1$), parity rules (even/odd), and divisibility tests.
Every valid cryptarithmetic puzzle is governed by strict mathematical axioms. Before attempting any deduction, confirm whether the puzzle adheres to standard alphametic conventions or supplies explicit custom constraints.
Each letter represents exactly one decimal digit from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. A letter cannot represent a negative number, a fraction, or a two-digit value.
Whenever a letter appears multiple times across different words or columns, it represents the exact same digit in every single occurrence throughout the entire puzzle.
Under standard alphametic conventions, different letters must represent different digits: if $A \ne B$, then $\text{digit}(A) \ne \text{digit}(B)$. No two distinct letters can share a numerical value.
The leftmost letter of any multi-digit word can never be assigned zero ($0$). For example, in $SEND$, $S \in \{1, 2, \dots, 9\}$. Single-digit numbers may be zero only if explicitly permitted.
The arithmetic operation must hold strictly in every column, taking into account the incoming carry or borrow from the right and the outgoing carry or borrow to the left.
A carry or borrow cannot appear or disappear arbitrarily. In standard two-number addition, the carry $k$ between adjacent columns is strictly binary: $k \in \{0, 1\}$.
A proposed mapping must satisfy the complete numerical equation simultaneously. An assignment that works in the units and tens columns but fails in the thousands column is completely invalid.
A puzzle may mathematically have zero solutions (inconsistent constraints), exactly one solution (unique), or multiple solutions. Standard exam puzzles are constructed to have a proven unique assignment.
Cryptarithmetic is not limited to addition. MBA exams test six distinct structural formats, each requiring a tailored algebraic approach:
Addition Puzzles
Two or more words added together to produce a sum word ($SEND + MORE = MONEY$).
Subtraction Puzzles
One word subtracted from another to yield a difference ($APPLE - THAT = EAT$).
Multiplication Puzzles
A multi-digit number multiplied by a single digit or word ($ABCD \times 9 = DCBA$).
Division Puzzles
Long division grids where quotient, divisor, dividend, and intermediate subtractions share letter encodings.
Missing-Digit Arithmetic
Mixed numerical grids where asterisks ($*$) or isolated letters conceal unknown digits in an otherwise known equation.
Multi-Equation Systems
Two or more simultaneous arithmetic equations sharing overlapping letter variables ($AB + BA = CDC$ and $BA - AB = 45$).
The core engine of cryptarithmetic is column algebra. Rather than looking at entire words, break the problem into modular single-column equations from right to left.
For any column with addend digits $a$ and $b$, result digit $c$, and incoming carry $k_{\text{in}} \in \{0, 1\}$, the outgoing carry $k_{\text{out}} \in \{0, 1\}$.
➕ Addition Carry & Column Mechanics
-
1.
Units Column: No incoming carry ($k_{\text{in}} = 0$). Hence $a + b = c + 10 \cdot k_{\text{out}}$. If $a + b = a$, then $b$ must be $0$.
-
2.
Max Column Sum: With two addends, maximum sum is $9 + 9 + 1 = 19$. The outgoing carry $k_{\text{out}}$ can only be $0$ or $1$. It can never be $2$ or higher.
-
3.
Repeated Letters: If $a + a = c$, then $c$ must be even ($0, 2, 4, 6, 8$). If $a + a + 1 = c$, then $c$ must be odd.
-
4.
Leading Carry Overflow: When an $n$-digit number plus an $n$-digit number produces an $(n+1)$-digit number, the leading digit of the sum must be 1 ($k_{\text{out}} = 1$).
➖ Subtraction Borrow Mechanics
-
1.
Direct Subtraction: If $a \ge b + \text{borrow}_{\text{in}}$, then no borrow is needed from the left: $\text{borrow}_{\text{out}} = 0$, and $c = a - b - \text{borrow}_{\text{in}}$.
-
2.
Active Borrow: If $a < b + \text{borrow}_{\text{in}}$, 1 unit is borrowed from the left column (adding 10 to current column): $c = 10 + a - b - \text{borrow}_{\text{in}}$, setting $\text{borrow}_{\text{out}} = 1$.
-
3.
Borrow Across Zeros: When borrowing from a column with digit $0$, that digit temporarily becomes $10$, donates $1$ to the right, and remains as $9$ for its own column evaluation.
✖️➗ Multiplication & Division Modular Constraints
For multiplication and division, single-column carry addition is replaced by place-value modular arithmetic and quotient-remainder bounding:
Never guess blindly. Follow this repeatable 11-step master algorithm to crack any cryptarithmetic puzzle systematically:
Align addends, subtrahends, products, and result words by exact place-value columns (Units, Tens, Hundreds, Thousands) using monospace formatting.
Mark all initial letters of multi-digit numbers. Explicitly record that $Letter \ne 0$ ($Letter \in \{1, 2, \dots, 9\}$).
List all distinct letters in the puzzle. The count must be $\le 10$. If the count exceeds 10, the puzzle has no solution in base 10.
Look for sum words that have one more digit than the addends. The leading carry digit is unconditionally locked to 1.
Look for $X + Y = X \implies Y = 0$ (if no incoming carry) or $X + Y + 1 = X \implies Y = 9$ (with carry).
Write explicit linear equations for each column: $Col_0: a + b = c + 10k_1$; $Col_1: d + e + k_1 = f + 10k_2$.
As soon as a letter's digit is proven, substitute it into every single position that letter occupies across all columns.
Cross out assigned digits from $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ to monitor which candidate digits remain eligible.
If direct deduction bifurcates into 2 plausible branches, test Branch A. Follow column propagation until a contradiction emerges or the puzzle completes.
Perform the full forward calculation numerically. Ensure no column overflow mismatch, no duplicate digit reuse, and no leading zero violation.
Confirm that all discarded branches genuinely generated arithmetic contradictions, establishing that exactly one unique assignment exists.
Study these 6 fully worked, mathematically verified examples progressing from simple 2-digit additions to multi-equation systems. Every deduction is justified step-by-step with explicit column equations and uniqueness proofs.
The Puzzle: $TO + GO = OUT$
- Hundreds Column (Leading Carry): The sum of two 2-digit numbers produces a 3-digit number $OUT$. The maximum sum of two 2-digit numbers is $99 + 99 = 198$. Therefore, the leading overflow digit $O = 1$.
- Propagate $O = 1$: Substitute $O = 1$ across all occurrences. The puzzle becomes: T1 + G1 = 1UT
- Units Column: $O + O = T \implies 1 + 1 = T \implies \mathbf{T = 2}$. Since $1 + 1 = 2 < 10$, there is no outgoing carry ($k_1 = 0$) to the tens column.
- Tens Column: $T + G = 10 \cdot O + U \implies 2 + G = 10(1) + U \implies 2 + G = 10 + U$.
- Since $G$ is a single digit ($\le 9$), $2 + G \le 11$. For $2 + G$ to reach at least $10$, $G$ must be $8$ or $9$.
- If $G = 9$: $2 + 9 = 11 \implies U = 1$. But $O = 1$, violating the distinct-letter rule ($U \ne O$). Thus $G \ne 9$.
- Hence $G$ must be $8$: $2 + 8 = 10 \implies \mathbf{G = 8}$ and $\mathbf{U = 0}$.
| Letter | T | G | O | U |
|---|---|---|---|---|
| Digit | 2 | 8 | 1 | 0 |
Units: $1 + 1 = 2$ (carry 0). Tens: $2 + 8 = 10$ (digit 0, carry 1). Hundreds: carry 1. Equation is 100% verified.
The Puzzle: $SEND + MORE = MONEY$
- Ten-Thousands Column: Sum of two 4-digit numbers produces a 5-digit number. The maximum possible sum is $9999 + 9999 = 19998$. Hence the leading carry $M = 1$.
- Thousands Column: $S + M + k_3 = O + 10 \cdot M \implies S + 1 + k_3 = O + 10$ (where $k_3 \in \{0, 1\}$ is carry from hundreds). To generate carry $1$, $S + 1 + k_3 \ge 10 \implies S + k_3 \ge 9$. If $S = 8$, $k_3$ must be $1$, giving $8 + 1 + 1 = 10 \implies O = 0$. If $S = 9$, $9 + 1 + k_3 = 10 + k_3 \implies O = k_3$. Since $M = 1$, if $O = k_3 = 1$, $O = M$ (conflict). Thus $O$ cannot be $1$, forcing $O = 0$ and $S = 9$ (with $k_3 = 0$ or $k_3 = 1$).
- Hundreds Column: $E + O + k_2 = N + 10k_3 \implies E + 0 + k_2 = N + 10k_3$. Since $E \ne N$, $k_2$ cannot be $0$ if $k_3 = 0$. Thus there must be a carry $k_3 = 0$ with $k_2 = 1$, giving $E + 1 = N$. (Note: $N$ is strictly one greater than $E$).
- Tens Column: $N + R + k_1 = E + 10 \cdot k_2 = E + 10$ (since $k_2 = 1$). Substitute $N = E + 1$: $(E + 1) + R + k_1 = E + 10 \implies R + k_1 + 1 = 10 \implies R + k_1 = 9$. Since distinct digits require $R \ne 9$ ($S = 9$), $k_1$ must be $1$, which forces $R = 8$.
- Units Column: $D + E = Y + 10 \cdot k_1 = Y + 10$. Because $k_1 = 1$, $D + E \ge 12$. Available digits are $\{2, 3, 4, 5, 6, 7\}$ (since $0, 1, 8, 9$ are used). Since $N = E + 1$, $E$ cannot be $7$ (would make $N = 8 = R$) or $6$ (would make $N = 7$; then $D+E \ge 12$ requires $D \ge 6$, but $6, 7, 8, 9$ all used). Testing $E = 5 \implies \mathbf{N = 6}$. Then $D + 5 = Y + 10 \implies D - Y = 5$. From remaining digits $\{2, 3, 4, 7\}$, the only pair with difference 5 is $D = 7$ and $Y = 2$. Hence $D = 7$ and $Y = 2$.
| S | E | N | D | M | O | R | Y |
|---|---|---|---|---|---|---|---|
| 9 | 5 | 6 | 7 | 1 | 0 | 8 | 2 |
Units: $7 + 5 = 12$ ($Y=2$, carry 1). Tens: $6 + 8 + 1 = 15$ ($E=5$, carry 1). Hundreds: $5 + 0 + 1 = 6$ ($N=6$, carry 0). Thousands: $9 + 1 = 10$ ($O=0$, carry 1). Ten-thousands: 1 ($M=1$). Exact match.
The Puzzle: $BASE + BALL = GAMES$
- Ten-Thousands (Leading Carry): Sum of two 4-digit numbers creates a 5-digit number $\implies \mathbf{G = 1}$.
- Thousands Column: $B + B + k_3 = 10 \cdot G + A = 10 + A$. Since $2B + k_3 \ge 10$, $B$ must be $\ge 5$. Also, $B \ne G \implies B \ne 1$.
- Hundreds Column: $A + A + k_2 = M + 10k_3 \implies 2A + k_2 = M + 10k_3$. Look at thousands column: $2B + k_3 = 10 + A$. Because $2B$ is even, the parity of $A$ equals the parity of $k_3$.
- Through systematic column propagation, testing $B = 7$: If $B = 7$, $2(7) + k_3 = 14 + k_3$. For this to be $10 + A$, $k_3$ must be $0$, which gives $\mathbf{A = 4}$.
- With $A = 4$ and $k_3 = 0$: Hundreds column requires $2(4) + k_2 = M \implies 8 + k_2 = M$. Since $M \le 9$ and $M \ne A, B, G$, $k_2$ must be $1$, forcing $\mathbf{M = 9}$.
- Tens Column with $k_2 = 1$: $S + L + k_1 = E + 10$. Units Column: $E + L = S + 10k_1$. Adding tens and units: $(S + L + k_1) + (E + L) = (E + 10) + (S + 10k_1) \implies 2L = 10 + 9k_1$. Since $2L$ must be an even integer, $k_1$ must be $0$, yielding $2L = 10 \implies \mathbf{L = 5}$.
- With $L = 5$ and $k_1 = 0$: Units column becomes $E + 5 = S$. Remaining unused digits from $\{0, 2, 3, 6, 8\}$: The only pair with difference 5 is $E = 3$ and $S = 8$. Hence $\mathbf{E = 3}$ and $\mathbf{S = 8}$.
| B | A | S | E | L | G | M |
|---|---|---|---|---|---|---|
| 7 | 4 | 8 | 3 | 5 | 1 | 9 |
Units: $3 + 5 = 8$ (carry 0). Tens: $8 + 5 = 13$ (digit 3, carry 1). Hundreds: $4 + 4 + 1 = 9$ (carry 0). Thousands: $7 + 7 = 14$ (digit 4, carry 1). Ten-thousands: 1. Completely verified.
The Puzzle: $APPLE - THAT = EAT$
- Ten-Thousands & Thousands Columns: $APPLE$ is a 5-digit number, but subtracting 4-digit $THAT$ leaves only a 3-digit number $EAT$. This requires $APPLE - THAT < 1000$. For a 5-digit number to be within 1000 of a 4-digit number (which is $\le 9999$), $APPLE$ must be barely above 10,000. This uniquely forces $A = 1$ and $P = 0$, making $APPLE = 100LE$.
- Also, since $100LE - THAT < 1000$, $THAT$ must be in the 9000s, which uniquely forces $T = 9$. Now our known assignments are: $A = 1, P = 0, T = 9$.
- Units Column Borrowing: $E - T = T \implies E - 9 = 9$. Since $E$ is a single digit, it must borrow $1$ (worth 10) from column $L$: $(10 + E) - 9 = 9 \implies 10 + E = 18 \implies \mathbf{E = 8}$.
- Tens Column Borrowing: Because the units column borrowed 1 from $L$, the tens column effective minuend is $L - 1$. $(L - 1) - A = A \implies (L - 1) - 1 = 1 \implies L - 2 = 1 \implies \mathbf{L = 3}$ (no borrow needed from hundreds).
- Hundreds Column Borrowing: $P - H = E \implies 0 - H = 8$. Since $0 < H$, it borrows $1$ from the thousands column (which is $P = 0$, propagating borrow from ten-thousands $A = 1$): $(10 + 0) - H = 8 \implies 10 - H = 8 \implies \mathbf{H = 2}$.
- Thousands & Ten-Thousands Verification: Thousands: $P=0$ became $10$ after borrowing from $A=1$, then lent $1$ to hundreds, leaving $9$. $9 - T = 9 - 9 = 0$. Ten-thousands: $A=1$ lent $1$, leaving $0$. Both columns cancel out cleanly!
| A | P | L | E | T | H |
|---|---|---|---|---|---|
| 1 | 0 | 3 | 8 | 9 | 2 |
Units: $18 - 9 = 9$. Tens: $(3-1) - 1 = 1$. Hundreds: $10 - 2 = 8$. Thousands: $(10-1) - 9 = 0$. Ten-thousands: $1-1=0$. Exact numerical verification.
The Puzzle: $ABCD \times 9 = DCBA$
- Thousands Magnitude Bound: For $ABCD \times 9$ to remain a 4-digit number ($< 10,000$), $A$ must satisfy: $A \times 9 < 10 \implies A \le 1$. Since $A \ne 0$ (leading digit), $A = 1$.
- Units Modular Product: The units calculation is $D \times 9 = \dots A \implies D \times 9$ must end in $1$. Testing digits $0$ to $9$ multiplied by $9$: only $9 \times 9 = 81$ ends in $1$. Therefore, $D = 9$ with an outgoing carry of $k_1 = 8$ to the tens column.
- Hundreds Column Carry Bound: Look at the thousands column of the product: $A \times 9 + k_3 = D \implies 1 \times 9 + k_3 = 9 \implies k_3 = 0$. There is zero carry from the hundreds column into the thousands column!
- For $B \times 9 + k_2$ to generate zero carry ($k_3 = 0$), $B \times 9 + k_2 < 10$. Since $B$ cannot be $1$ ($A = 1$), $B$ must be $0$. (If $B \ge 2$, $2 \times 9 = 18 > 10$, generating a carry). Thus $B = 0$.
- Tens Column Modular Equation: The tens column calculation is: $C \times 9 + k_1 = \dots B \implies C \times 9 + 8 = \dots 0$. This means $C \times 9$ must end in $2$ (since $2 + 8 = 10$). The unique single-digit multiple of $9$ ending in $2$ is $8 \times 9 = 72$. Therefore, $C = 8$. Let's check the carry: $8 \times 9 + 8 = 80$, so the digit is $0$ ($B=0$) and carry $k_2 = 8$.
- Check Hundreds with $k_2 = 8$: $B \times 9 + k_2 = 0 \times 9 + 8 = 8 = C$. The hundreds digit is exactly $C = 8$, and the carry into thousands is $0$ ($k_3 = 0$). Everything matches with 100% precision!
| A | B | C | D |
|---|---|---|---|
| 1 | 0 | 8 | 9 |
Notice that $1089$ reversed is exactly $9801$. All 4 digits $\{0, 1, 8, 9\}$ are distinct. Equation holds unconditionally.
The Puzzle: Coupled Cryptarithmetic System
- Algebraic Expansion of Equation 1: $AB = 10A + B$ and $BA = 10B + A$. $$AB + BA = (10A + B) + (10B + A) = 11(A + B)$$ Thus, the 3-digit palindrome $CDC$ must be a multiple of 11.
- Divisibility Rule of 11 on $CDC$: For a 3-digit number $CDC$ to be divisible by 11, the alternating sum of digits must be a multiple of 11: $(C + C) - D = 2C - D \in \{0, 11\}$. Since $AB$ and $BA$ are 2-digit numbers, their sum $11(A + B)$ is at most $11(9 + 8) = 187$. Because $CDC$ is a 3-digit number starting with $C$, $C$ can only be $1$ ($CDC$ is in the 100s). Hence $C = 1$.
- With $C = 1$: $2(1) - D = 0 \implies \mathbf{D = 2}$. Therefore, $CDC = 121$. Since $11(A + B) = 121 \implies \mathbf{A + B = 11}$.
- Algebraic Expansion of Equation 2: $$BA - AB = (10B + A) - (10A + B) = 9(B - A)$$ Given $BA - AB = 45 \implies 9(B - A) = 45 \implies \mathbf{B - A = 5}$.
- Solve the Linear System: We have two simultaneous linear equations in single digits: $$\begin{cases} A + B = 11 \\ B - A = 5 \end{cases}$$ Adding both equations: $2B = 16 \implies \mathbf{B = 8}$. Substituting $B = 8$: $A + 8 = 11 \implies \mathbf{A = 3}$.
- Check Distinctness: The digits are $A = 3, B = 8, C = 1, D = 2$. All 4 digits $\{1, 2, 3, 8\}$ are completely distinct! None are zero.
| A | B | C | D |
|---|---|---|---|
| 3 | 8 | 1 | 2 |
Both equations are satisfied simultaneously without contradiction.
Assigning $0$ to an initial letter of a multi-digit number (e.g. letting $S=0$ in $SEND$) invalidates the solution under standard decimal notation rules.
Assigning one digit to a letter in the units column and a different digit to the same letter in the tens column breaks the fundamental uniformity axiom.
Under standard all-different rules, once digit $7$ is assigned to $D$, no other letter ($E, N, R$, etc.) can ever be $7$.
Calculating $a + b = c$ without checking if the previous column produced an incoming carry $k_{\text{in}} = 1$ is the most frequent calculation error.
Fixing a value for letter $E$ based only on the units column without verifying that the same value works in the tens and hundreds columns where $E$ also appears.
Validating individual columns in isolation without checking whether the entire number sum balances as a single holistic equation.
Treating a working guess as proven fact without demonstrating that competing candidate digits produce mathematical contradictions.
Assuming that finding one valid mapping guarantees uniqueness. In puzzles with free variables, multiple distinct mappings can exist.
Plugging random digits into unconstrained columns instead of anchoring deductions to leading carries, zero anchors, or parity rules.
In multi-equation sets, solving Equation 1 without verifying that the assigned values satisfy Equation 2 and Equation 3.
Whenever two $n$-digit numbers sum to an $(n+1)$-digit number, the leading digit of the sum must be 1 ($k_{\text{out}} = 1$). If three $n$-digit numbers sum to $(n+1)$ digits, the leading digit is $\le 2$.
In $SEND + MORE = MONEY$, for $S + M + k = MO$, since $M = 1$, $S + 1 + k \ge 10$. Because $O \ne M$ ($O \ne 1$), $O$ is forced to $0$ and $S$ is forced to $9$.
$A + A = \text{Even}$. Therefore, if $A + A = C$ without carry, $C \in \{0, 2, 4, 6, 8\}$. If $A + A + 1 = C$, $C$ must be odd. Parity immediately eliminates half of all digit candidates.
Check digital roots: $\text{Sum}(\text{Addend}_1) + \text{Sum}(\text{Addend}_2) \equiv \text{Sum}(\text{Result}) \pmod 9$. This catches arithmetic slips in seconds without re-solving individual columns.
In $WORD \times N = RESULT$, if both are 4-digit words, $W \times N < 10$. If $N = 4$, $W \in \{1, 2\}$. If $W$ is known to be even, $W = 2$ immediately.
When column deductions leave only 2 choices for a letter, do not guess. Test the more restrictive value. If it causes an illegal duplicate within 2 steps, immediately adopt the alternative.
In time-pressured competitive exams (CAT, SNAP, MBA CET), candidates who treat cryptarithmetic as a sudoku or trial-and-error game lose 10 to 15 minutes per set. To maximize efficiency:
Test your mastery across all operations (addition, subtraction, multiplication, division, missing-digit, and coupled systems). Click "Show Solution" on each question to inspect the verified step-by-step mathematical deduction.
+ M E
B E E
Step-by-Step Deduction:
- Units Column: $E + E = E$ (or $10 + E$). If $2E = E \implies E = 0$. If $2E = 10 + E \implies E = 10$ (not a single digit). Thus $E = 0$, with no carry to tens ($k_1 = 0$).
- Tens & Hundreds: $M + M = BE \implies 2M = 10B + E = 10B + 0 = 10B$. Thus $2M = 10B \implies M = 5B$.
- Since $M$ is a single digit ($\le 9$) and $B \ne 0$, the only possible integer value for $B$ is $B = 1$, which gives $M = 5$.
- Verification: $50 + 50 = 100$. All letters $\{M=5, E=0, B=1\}$ are distinct. Sum $= 5 + 0 + 1 = 6$. Proven 100% unique.
Step-by-Step Deduction:
- $HE \times HE = SHE$ means the square of a 2-digit number ending in $E$ produces a 3-digit number ending in the same two digits $HE$. An integer whose square ends in the same digits is an automorphic number.
- Units digit $E^2$ must end in $E$. Single digits with this property are $0, 1, 5, 6$.
- If $E = 0$: $H0^2 = H^2 00$. The tens digit would be $0 \ne H$ (since $H \ne 0$).
- If $E = 1$: $(10H + 1)^2 = 100H^2 + 20H + 1$. Tens digit is $(2H) \pmod{10}$. For $2H \equiv H \pmod{10}$, $H = 0$ (impossible).
- If $E = 5$: Any number ending in 25 has its square ending in 25 ($25^2 = 625$). Here $HE = 25 \implies H = 2, E = 5$, and $SHE = 625 \implies \mathbf{S = 6}$.
- Checking $E = 6$: Numbers ending in 76 have square ending in 76 ($76^2 = 5776$, 4 digits, not 3).
- Thus the unique solution is $25 \times 25 = 625$. Letters: $H = 2, E = 5, S = 6$ (all distinct). Therefore, $\mathbf{S = 6}$.
+ T W O
F O U R
Step-by-Step Deduction:
- Given $O = 4$. Leading carry from 3-digit sum to 4-digit sum forces $F = 1$.
- Units Column: $O + O = R \implies 4 + 4 = 8 \implies \mathbf{R = 8}$, with no carry ($k_1 = 0$).
- Hundreds Column: $T + T + k_2 = 10F + O = 10(1) + 4 = 14$. If $k_2 = 1$, $2T + 1 = 14 \implies 2T = 13$ (impossible). Hence $k_2 = 0$, giving $2T = 14 \implies \mathbf{T = 7}$.
- Tens Column: Because $k_2 = 0$, $W + W < 10 \implies 2W = U$ (no carry to hundreds). Available unused digits: $\{0, 2, 3, 5, 6, 9\}$ (since $F=1, O=4, T=7, R=8$ are used). If $W = 2 \implies U = 4$ (already used by $O$). If $W = 3 \implies \mathbf{U = 6}$ (both 3 and 6 are available!). If $W = 5 \implies U = 10$ (invalid).
- Thus uniquely $W = 3$ and $U = 6$. Verification: $734 + 734 = 1468$. Solution is 100% unique.
+ T E N
+ T E N
S I X T Y
Step-by-Step Deduction:
- Units Column: $Y + N + N = Y + 10k_1 \implies 2N = 10k_1$. Since $N$ cannot equal $Y$ and $2N$ must be a multiple of 10, $N$ can only be $0$ or $5$. If $N = 5$, $2N = 10 \implies k_1 = 1$. If $N = 0$, $k_1 = 0$. Looking at tens column: $T + E + E + k_1 = T + 10k_2 \implies 2E + k_1 = 10k_2$. This forces $2E + k_1$ to be a multiple of 10. Since $2E$ is even, $k_1$ must be even $\implies \mathbf{k_1 = 0}$, which uniquely establishes $N = 0$.
- Because $k_1 = 0$, $2E = 10k_2$. Since $E \ne N$ ($E \ne 0$), $E$ must be $E = 5$, and $k_2 = 1$.
- Complete Assignment: Continuing column algebra gives $F = 2, O = 9, R = 7, T = 8, Y = 6, E = 5, N = 0, S = 3, I = 1, X = 4$. Numerical equation: $29786 + 850 + 850 = 31486$. Every digit 0 to 9 is used exactly once. Thus $\mathbf{E = 5}$.
+ R O A D S
D A N G E R
Step-by-Step Deduction:
- Hundred-Thousands (Leading Carry): Adding two 5-digit numbers produces a 6-digit number $DANGER$. The leading overflow digit must be 1: $\mathbf{D = 1}$.
- Ten-Thousands Column: $C + R + k_4 = 10 \cdot D + A = 10 + A$. To generate a carry of 10, $C + R \ge 9$.
- Units Column: $S + S = R + 10k_1 \implies 2S = R + 10k_1$. This proves $R$ must be an even digit ($R \in \{2, 4, 6, 8\}$).
- Through column parity and carry constraints: If $R = 6$, $2S = 6 + 10k_1 \implies$ if $k_1 = 0, S = 3$. Tens column: $S + D + k_1 = E \implies 3 + 1 + 0 = 4 \implies E = 4$ (no carry to hundreds, $k_2 = 0$). Hundreds column: $O + A = N + 10k_3$. Thousands column: $R + O + k_3 = A + 10k_4 \implies 6 + O + k_3 = A + 10k_4$. Testing remaining digits yields uniquely $C = 9, R = 6, O = 2, S = 3, A = 5, D = 1, N = 8, G = 7, E = 4$.
- Verification: $96233 + 62513 = 158746$. Distinct digits verified. $\mathbf{D = 1}$.
- B A L L
B A S E
Step-by-Step Deduction:
- Invert the subtraction into equivalent addition: $BALL + BASE = GAMES$.
- Notice this is structurally identical to Solved Example 3!
- Leading carry from 4-digit sum to 5-digit sum forces $G = 1$.
- Column algebra uniquely resolves the letters: $B = 7, A = 4, S = 8, E = 3, L = 5, G = 1, M = 9$.
- Check Subtraction: $14938 - 7455 = 7483$. Units: $8 - 5 = 3$ ($E=3$). Tens: $13 - 5 = 8$ (borrow 1, $S=8$). Hundreds: $(9-1) - 4 = 4$ ($A=4$). Thousands: $14 - 7 = 7$ ($B=7$). Ten-thousands: $1 - 1 = 0$.
- $G + A + M + E + S = 1 + 4 + 9 + 3 + 8 = 25$.
Step-by-Step Deduction:
- Thousands Bound: For $ABCD \times 4$ to produce a 4-digit number ($< 10,000$), $A \times 4 < 10 \implies A \in \{1, 2\}$. Also, the units digit of the product is $D \times 4 = \dots A$. Since any multiple of 4 is even, $A$ must be even. This forces $A = 2$.
- Units Column: $D \times 4$ must end in $A = 2$. Single digits whose product with 4 ends in 2 are $3$ and $8$ ($3 \times 4 = 12, 8 \times 4 = 32$). From thousands column: $A \times 4 = 2 \times 4 = 8 \le D$. Thus $D \ge 8$. This uniquely forces $D = 8$. Carry to tens is $k_1 = 3$ ($8 \times 4 = 32$).
- Thousands Carry: $A \times 4 + k_3 = D \implies 2 \times 4 + k_3 = 8 \implies k_3 = 0$. No carry from hundreds!
- For $B \times 4 + k_2$ to generate zero carry ($k_3 = 0$), $B \times 4 + k_2 < 10 \implies B \in \{0, 1, 2\}$. Since $A = 2$, $B \ne 2$. If $B = 0$: tens column $C \times 4 + 3 = \dots B = \dots 0$, but an even number plus 3 is odd (impossible). Hence $B$ must be $B = 1$.
- Tens Column: $C \times 4 + 3 = \dots 1 \implies C \times 4$ ends in $8$. Candidate digits for $C$ are $2$ and $7$. Since $A = 2$, $C$ must be $C = 7$. Check hundreds: $1 \times 4 + \text{carry } 3 = 7 = C$. Exactly matches!
- Verification: $2178 \times 4 = 8712$. Reversed order verified. $\mathbf{ABCD = 2178}$.
Step-by-Step Deduction:
- Rewrite division as multiplication: $C \times (10D + 4) = 700 + 10A + B$.
- The product is in the 700s ($700 \le \text{Product} \le 799$). Since $C$ is an even single-digit divisor, $C \in \{2, 4, 6, 8\}$. Notice that digit $4$ is already used, so $C \in \{2, 6, 8\}$.
- If $C = 2$: $2 \times (10D + 4) \le 2 \times 94 = 188 < 700$ (too small).
- If $C = 6$: $6 \times (10D + 4) \ge 700 \implies 10D + 4 \ge 117$, impossible since $D4$ is a 2-digit number ($\le 94$).
- Thus $C$ must be $C = 8$.
- With $C = 8$: $8 \times (10D + 4) \in [700, 799] \implies 10D + 4 \ge 87.5 \implies D \ge 9$. Since $D$ is a single digit, $D = 9$.
- Calculate product: $8 \times 94 = 752$. Comparing with $7AB$: $\mathbf{A = 5}$ and $\mathbf{B = 2}$.
- Check distinctness of $\{7, 4, A=5, B=2, C=8, D=9\}$: All 6 digits $\{2, 4, 5, 7, 8, 9\}$ are completely distinct. Quotient is $D4 = 94$.
+ B 3 8
7 9 C
Step-by-Step Deduction:
- Printed digits already present: $\{1, 2, 3, 7, 8, 9\}$. Available candidate digits for $\{A, B, C\}$ from the allowed set are $\{0, 5, 6\}$.
- Units Column: $2 + 8 = 10$. This immediately requires result digit $C = 0$ and generates a carry $k_1 = 1$ to the tens column.
- Tens Column: $A + 3 + k_1 = 9 \implies A + 3 + 1 = 9 \implies A + 4 = 9 \implies \mathbf{A = 5}$. Since $A + 4 = 9 < 10$, there is zero carry ($k_2 = 0$) into the hundreds column.
- Hundreds Column: $1 + B + k_2 = 7 \implies 1 + B + 0 = 7 \implies \mathbf{B = 6}$.
- Verification: $152 + 638 = 790$. Digits used: $\{1, 5, 2, 6, 3, 8, 7, 9, 0\}$. Every digit is distinct. Solution is unique.
(2) CD + EF = GHI
Step-by-Step Deduction:
- The 9 letters are a permutation of digits $1$ through $9$. Neither digit 0 nor any duplicate digit is permitted.
- Equation (2): $CD + EF = GHI$. Since $CD \le 98$ and $EF \le 98$, the sum $GHI \le 196$. Because 0 is not used, $G$ must be $G = 1$.
- Equation (1): $A \times B = CD$. Single-digit products using distinct digits without 0.
An exhaustive constraint solver shows that among all permutations of digits $1..9$, exactly two mathematical solutions exist:
- Branch 1: $6 \times 9 = 54$, and $54 + 78 = 132$ ($GHI = 132$, an even number).
- Branch 2: $7 \times 9 = 63$, and $63 + 82 = 145$ ($GHI = 145$, an odd number).
- The puzzle explicitly specifies that $GHI$ is an even number. This cleanly eliminates Branch 2 and isolates Branch 1 as the single unique solution!
- Letter assignments: $A = 6, B = 9, C = 5, D = 4, E = 7, F = 8, G = 1, H = 3, I = 2$. All 9 digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ appear exactly once.
- The question asks for $CD \times B = 54 \times 9 = \mathbf{486}$.
What is cryptarithmetic?
Cryptarithmetic (also called alphametics) is a mathematical puzzle where letters or symbols replace decimal digits in an arithmetic equation. The objective is to deduce the unique digit represented by each letter so that the mathematical operation remains valid.
Can two different letters represent the same digit?
In standard alphametic puzzles, no. Different letters must represent different digits from $0$ to $9$ (the injective rule). If duplicate assignments are permitted, the exam problem will explicitly state "letters need not represent distinct digits".
Can a leading letter be zero?
No. Standard decimal notation forbids multi-digit numbers from starting with zero ($0$). Thus, any initial letter of a multi-digit number must be an integer between $1$ and $9$.
How do carries work in cryptarithmetic?
When adding two single digits plus an incoming carry $k_{\text{in}} \le 1$, the maximum column sum is $9 + 9 + 1 = 19$. Therefore, the outgoing carry to the adjacent left column can only ever be $0$ or $1$. In particular, the leading overflow digit of a sum is always $1$.
How should subtraction and borrowing be handled?
In subtraction, when a column's minuend digit is smaller than the subtrahend digit, 1 unit (worth 10) is borrowed from the left. An effective shortcut is to invert the subtraction equation $A - B = C$ into the equivalent addition equation $B + C = A$ and apply standard carry rules.
Is trial and error necessary?
Blind trial and error is never necessary and wastes critical exam time. Solvers should apply boundary constraints, parity rules, and column equations. Controlled trial (evaluating at most 2 pruned branches) should only be used when direct deduction narrows a letter to two choices.
How do I prove that a solution is unique?
A solution is uniquely proven when every alternative branch has been shown to produce an arithmetic contradiction, duplicate a used digit, or violate the non-zero leading digit rule.
Can a cryptarithmetic puzzle have multiple solutions?
Yes, if the puzzle has free unconstrained variables or symmetrical letters (e.g. swapping $A$ and $B$ produces the same column sum). Standard exam questions formulate their prompt around invariant properties or provide sufficient constraints to guarantee a unique solution.
What should I do when a puzzle has too many possible assignments?
Target the most constrained position: look for the leading carry, evaluate column parity (even/odd), check divisibility by 9 across the words, and use the multiple-choice options to eliminate improbable values.
Cryptarithmetic bridges algebraic substitution and analytical puzzle solving. Continue your structured LR preparation with these adjacent concepts:
Advance to Syllogism — master 2 and 3-statement Venn diagrams, the 100-50 tick-cross methodology, All/Some/No conversions, and possibility conclusion rules for CAT, SNAP, and MBA CET.