Cryptarithmetic Problems: Rules, Techniques, and 3 Solved Examples

Cryptarithmetic problems replace every digit in an arithmetic equation with a letter; your job is to find the unique digit assignment that makes the arithmetic hold.

The puzzles appear in placement aptitude rounds at TCS, Infosys, Wipro, and Cognizant as 1 to 3 questions testing systematic reasoning rather than memorised formulas. A student who has never seen one before will stare at SEND + MORE = MONEY and feel stuck. A student who knows the six governing rules and five solving techniques will crack that same puzzle in under 8 minutes.

What Is a Cryptarithmetic Problem

A cryptarithmetic problem (also called verbal arithmetic or an alphametic) is an arithmetic equation written entirely in letters, where each letter stands for a distinct decimal digit. The term was coined by puzzle-maker Henry Dudeney in the early twentieth century. The most-studied example is the addition SEND + MORE = MONEY, first published in 1924.

Two properties define every valid cryptarithmetic puzzle:

• Unique mapping: each letter is assigned exactly one digit, and each digit is assigned to at most one letter across the entire puzzle.
• Unique solution: a well-formed puzzle has exactly one valid digit assignment. If two assignments both satisfy the equation, the puzzle is considered ambiguous and poorly constructed.

The terms “cryptarithmetic,” “verbal arithmetic,” and “alphametics” are used interchangeably in Indian placement prep material. All three refer to the same class of puzzle.

The 6 Rules Every Solver Must Know

Before assigning a single digit, internalise these rules. They eliminate most wrong assignments before you try them:

• Rule 1 — One letter, one digit. Every occurrence of the same letter maps to the same digit. If S=9 in one column, S=9 in every column.
• Rule 2 — One digit, one letter. Two different letters cannot share the same digit. If M=1, no other letter is 1.
• Rule 3 — No leading zeros. The first letter of any number in the puzzle cannot be 0. In SEND + MORE = MONEY, none of S, M (for MORE), or M (for MONEY) can be 0.
• Rule 4 — Base 10. All arithmetic is decimal unless the problem explicitly states a different base.
• Rule 5 — Carry is at most 1. In base-10 addition, the carry generated from any column into the next column is either 0 or 1. This is the single most powerful constraint in the puzzle.
• Rule 6 — Unique solution. A valid puzzle has exactly one correct assignment. If your analysis produces two viable branches, keep working: one will generate a contradiction.

Solving Techniques: 5 Methods That Work

1. Most-Significant-Column First

When a sum produces a result with more digits than the addends, the carry out of the top column equals the leading digit of the result. That carry can only be 0 or 1, and since the leading digit cannot be 0 (Rule 3), it must be 1. This single deduction immediately pins the leading letter’s value before you look at any other column.

2. Column-by-Column Constraint Propagation

Set up every column as an equation: the sum of the column’s digits plus any carry-in equals the result digit plus 10 times the carry-out. Write the unknowns and constraints, then propagate. Each pinned digit reduces the search space for remaining columns. Work right to left (units first) or left to right (most significant first), whichever gives you more immediate constraints.

3. High-Frequency Letter Analysis

A letter that appears in multiple positions constrains several columns simultaneously. In EAT + THAT = APPLE, the letter T appears in three positions. Identifying all the column equations that contain T and solving them together is far faster than working through each column in isolation.

4. Parity and Last-Digit Analysis

The units column has no carry-in. The sum of the units digits of the addends equals either the units digit of the result (carry-out = 0) or the units digit of the result plus 10 (carry-out = 1). The parity (odd or even) of the sum often pins the units digit of the result directly, which then cascades into the tens column.

5. Systematic Backtracking

After applying the first four techniques, you may have 2 to 3 candidate values for a key letter. Assume one value, propagate its consequences through all columns, and watch for contradictions (two letters forced to the same digit, or a digit forced outside 0 to 9). If a contradiction appears, backtrack and try the next candidate. Practiced solvers need at most 2 to 3 backtrack steps on a standard 8-letter puzzle.

Worked Examples with Full Solutions

SEND + MORE = MONEY

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

• Step 1 (leading digit): MONEY is a 5-digit number; SEND and MORE are each 4-digit numbers. The maximum sum of two 4-digit numbers is 9999 + 9999 = 19998, which always starts with 1. So M = 1.
• Step 2 (thousands column): S + M + carry3 = O + 10, because the ten-thousands carry = M = 1. With M=1: S = O + 9 - carry3. If carry3=0, then O=0 and S=9. If carry3=1, then S = O + 8, giving either (S=8, O=0) or (S=9, O=1). But O=1 would clash with M=1. Test S=9, O=0, carry3=0 first.
• Step 3 (hundreds column): E + O + carry2 = N + 10 x carry3 = N + 0. With O=0: E + carry2 = N. Since N and E must be different, carry2 = 1 and N = E + 1.
• Step 4 (tens column): N + R + carry1 = E + 10 x carry2 = E + 10. With N = E + 1: (E+1) + R + carry1 = E + 10, so R + carry1 = 9. If carry1=0 then R=9, but S=9 already. So carry1=1 and R=8.
• Step 5 (units column): D + E = Y + 10 x carry1 = Y + 10. Since carry1=1, we need D + E >= 10 and Y = D + E - 10. Digits used so far: M=1, O=0, S=9, R=8. Remaining digits for E, N (=E+1), D, Y come from 7. Try E=5, N=6: from the remaining set 7, take D=7, giving Y = 7 + 5 - 10 = 2. All eight digits are unique.
• Final assignment: S=9, E=5, N=6, D=7, M=1, O=0, R=8, Y=2. Verification: 9567 + 1085 = 10652. MONEY = 10652. Correct.

BASE + BALL = GAMES

  B A S E
+ B A L L
---------
G A M E S

• Step 1 (leading digit): GAMES has 5 digits; both addends have 4. Maximum sum is 19998, so G = 1.
• Step 2 (thousands column): B + B + carry3 = A + 10 (since G = carry4 = 1). So 2B + carry3 = A + 10.
  • carry3=0: A = 2B - 10. Valid pairs: B=5 gives A=0; B=6 gives A=2; B=7 gives A=4.
  • carry3=1: A = 2B - 9. Valid pairs: B=6 gives A=3; B=7 gives A=5; B=8 gives A=7.
• Step 3 (hundreds column, testing B=7, A=4, carry3=0): A + A + carry2 = M + 10 x carry3 = M + 0. So 2x4 + carry2 = M. With carry2=1: M=9. With carry2=0: M=8.
• Step 4 (tens and units, B=7, A=4, M=9, carry2=1): Tens: S + L + carry1 = E + 10. Units: E + L = S + 10 x carry1. Adding both equations and simplifying: 2L + carry1 = 10 x carry1 + 10. With carry1=0: 2L=10, so L=5. With carry1=1: 2L=19, which has no integer solution. So carry1=0 and L=5.
• Step 5 (resolving E and S): With carry1=0, units: E + 5 = S (no carry). With carry1=0, tens: S + 5 = E + 10, so S = E + 5. Both equations agree on S = E + 5. Digits used: G=1, B=7, A=4, M=9, L=5. Remaining from 8: take E=3, S=8. All digits unique.
• Final assignment: G=1, A=4, M=9, E=3, S=8, B=7, L=5. Verification: BASE=7483, BALL=7455, 7483 + 7455 = 14938. GAMES = 14938. Correct.

EAT + THAT = APPLE

    E A T
+  T H A T
----------
 A P P L E

• Step 1 (leading digit): APPLE has 5 digits; EAT has 3 and THAT has 4. The maximum sum is 999 + 9999 = 10998, which has at most 5 digits with a leading 1. Since the leading digit A cannot be 0, A = 1 and the carry out of the thousands column is 1.
• Step 2 (thousands column): T + carry3 = P + 10 x carry4 = P + 10. For P to be a valid digit (0 to 9), T + carry3 must equal exactly 10. Since T is at most 9 and carry3 is at most 1, the only solution is T=9 and carry3=1, giving P = 0.
• Step 3 (units column): T + T = E + 10 x carry1. With T=9: 18 = E + 10 x carry1. The only valid solution is carry1=1 and E=8.
• Step 4 (tens column): A + A + carry1 = L + 10 x carry2. With A=1 and carry1=1: 3 = L + 10 x carry2. Since L must be a single digit and non-negative, carry2=0 and L=3.
• Step 5 (hundreds column): E + H + carry2 = P + 10 x carry3. With E=8, carry2=0, P=0, carry3=1: 8 + H + 0 = 0 + 10, so H = 2.
• Final assignment: A=1, T=9, P=0, E=8, L=3, H=2. Verification: EAT=819, THAT=9219, 819 + 9219 = 10038. APPLE = 10038. Correct.

Note: P=0 is valid here because P is not the leading digit of APPLE (A=1 is the leading digit). Both P positions in APPLE hold the value 0.

Common Puzzle Formats in Placement Tests

The three worked examples above are all addition puzzles with a single carry direction. Placement tests sometimes introduce variations:

Addition with three or more addends: Problems like EARTH + AIR + FIRE + WATER = NATURE add three or four numbers. The carry-out from each column can be 0, 1, 2, or 3 depending on how many addends are in that column. The solving approach is identical; only the carry range expands.

Subtraction puzzles: Less common but tested at some companies. In subtraction, borrow propagates right to left (like carry propagates left to right in addition). Set up each column as: minuend digit minus subtrahend digit minus borrow-in equals result digit plus 10 times borrow-out.

Multiplication puzzles: Rare in standard placement tests but sometimes seen in D.E. Shaw aptitude rounds. Each partial product generates its own set of column constraints. The approach is the same; the number of equations per column increases.

Mixed formats with shared letters across two equations: Some puzzles present two separate equations sharing letters, effectively doubling the constraint set. Treat each equation’s columns independently, then look for the assignment that satisfies all constraints simultaneously.

The logical reasoning sections that include cryptarithmetic often appear alongside number analogy patterns and data sufficiency questions. Students who have practised constraint propagation on cryptarithmetic typically find those adjacent question types easier to manage within the time limit.

Companies that run analytical aptitude rounds heavily, including analytics firms covered in the Mu Sigma interview process guide, include constraint-based puzzles precisely because they reveal whether a candidate organises information before attempting to solve.

Practice Plan: From Zero to Test-Ready

Weeks 1 to 2: Rule internalisation

Work through at least 10 short puzzles with 3 to 4 unique letters, addition only. Do not time yourself yet. The goal is correctness: can you apply all six rules without looking them up, and can you set up the column equations cleanly? Standard starting puzzles include ONE + ONE = TWO and AB + BA = CBC.

Week 3: Classic puzzles with timing

Solve SEND + MORE = MONEY, BASE + BALL = GAMES, and EAT + THAT = APPLE from scratch (do not look at the solutions above). Time each attempt. The target is under 8 minutes per puzzle from a cold start. If you exceed 12 minutes, the bottleneck is usually failing to propagate carry constraints early, not backtracking errors.

Week 4: Exam simulation

Set a 5-minute timer, take a single unseen puzzle, and commit to column-by-column analysis without guessing. In a multiple-choice format, the option that matches your first confidently-deduced letter assignment is almost always correct, even if you have not verified the full puzzle. Partial analysis is a legitimate exam strategy when time is tight.

What not to do: trial-and-error from digit 0 upward covers roughly 1.8 million combinations for an 8-letter puzzle (the eight-from-ten permutation count). Even fast arithmetic does not cover that in 8 minutes. The techniques above reduce the search space to 2 to 3 forced choices; trial-and-error is what you fall back on only when constraints alone do not close the puzzle, and even then only for the last 1 or 2 undetermined letters.

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The column-by-column constraint narrowing you practised above is a manual version of what constraint-satisfaction algorithms (and language models) run automatically. TinkerLLM lets you test that directly at ₹299: prompt an LLM to solve a fresh cryptarithmetic puzzle, trace where it propagates constraints correctly, and notice whether it backtracks cleanly or stalls at a contradiction.