Letters for Digits Puzzles
Letters for digits puzzles (also called cryptarithmetic) are problems where letters replace digits in arithmetic operations. Your task is to find which digit each letter represents.
Each letter stands for a unique digit (0-9), and the same letter always represents the same digit throughout the puzzle. These puzzles use logic, divisibility rules, and the general form of numbers to find the answer.
This is part of the NCERT Chapter "Playing with Numbers" for Class 8.
What is Letters for Digits Puzzles?
Definition: A letters-for-digits puzzle is a mathematical problem where digits in an addition, subtraction, or multiplication are replaced by letters. The goal is to find the digit each letter represents.
Rules:
- Each letter represents exactly one digit (0-9).
- Different letters represent different digits.
- The leading digit of any number cannot be 0.
- The arithmetic operation must be correct when letters are replaced by digits.
Methods
Strategy for solving:
- Look at the carry — in addition, the maximum carry from one column is 1 (since max digit + max digit + carry = 9 + 9 + 1 = 19).
- Start with columns that have the fewest possibilities.
- Use logic and elimination — try values and check consistency.
- Check the units column first — it has no incoming carry.
- Use the constraint that different letters = different digits.
Common patterns to look for:
- If A + A = A in units column (with no carry), then A = 0.
- If A + B = 10 + something, there is a carry of 1 to the next column.
- If a 2-digit number + 2-digit number = 3-digit number, the result's hundreds digit must be 1.
Solved Examples
Example 1: Example 1: Simple addition
Problem: Find A and B:
3A
+ B3
———
89
Solution:
- Units column: A + 3 = 9 (or 19 with carry). Try A + 3 = 9, so A = 6.
- Tens column: 3 + B = 8. So B = 5.
- Check: 36 + 53 = 89 ✓
Answer: A = 6, B = 5.
Example 2: Example 2: Addition with carry
Problem: Find A:
A7
+ 3A
———
111
Solution:
- Units: 7 + A. If this gives a units digit of 1, then 7 + A = 11 (carry 1). So A = 4.
- Tens: A + 3 + 1(carry) = 4 + 3 + 1 = 8. But we need tens digit = 1 and hundreds digit = 1.
- So tens column: A + 3 + 1 = 11 (with carry). A + 4 = 11, A = 7. But units gave A = 4. Contradiction.
- Retry units: 7 + A gives units digit 1. So 7 + A = 11, A = 4. Tens: 4 + 3 + 1 = 8. Result should be 111, but tens digit is 8, not 1. Does not work.
- Retry: 7 + A = 21? Not possible (max = 9+9=18). So 7 + A = 11, A = 4, carry 1. Tens: 4 + 3 + 1 = 8. Hundreds = 0 + carry = 0. Result = 081. Not 111.
- This puzzle may need re-examination. Let's try A = 7: units: 7+7=14, write 4 carry 1. Tens: 7+3+1=11, write 1 carry 1. Hundreds: 1. Result = 114, not 111.
- Try A = 4: 47 + 34 = 81 (not 111). Not matching.
Let us re-read as: A7 + 3A = 111. Try A=7: 77+37=114. Try A=6: 67+36=103. Try A=5: 57+35=92. None give 111.
Revised problem: The puzzle has no solution — this demonstrates the importance of checking every candidate.
Answer: No single-digit A satisfies this puzzle.
Example 3: Example 3: NCERT classic
Problem: Find A and B:
AB
+ BA
———
121
Solution:
- AB + BA = (10A + B) + (10B + A) = 11(A + B) = 121
- A + B = 121 ÷ 11 = 11
- A and B are single digits, A + B = 11, and A ≠ 0, B ≠ 0.
- Possible: (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2).
- Check: 29 + 92 = 121 ✓, 38 + 83 = 121 ✓, etc.
Answer: Multiple solutions: A + B = 11. For example, A = 2, B = 9 (or A = 3, B = 8, etc.).
Example 4: Example 4: Multiplication puzzle
Problem: Find A:
1A
× A
———
9A
Solution:
- (10 + A) × A = 90 + A
- 10A + A² = 90 + A
- A² + 9A − 90 = 0
- (A + 15)(A − 6) = 0
- A = 6 (rejecting A = −15)
- Check: 16 × 6 = 96. Written as 9A = 96. A = 6 ✓
Answer: A = 6.
Example 5: Example 5: Subtraction puzzle
Problem: Find A and B:
A8
− 2B
———
19
Solution:
- Units: 8 − B = 9. Since 8 < 9, borrow: 18 − B = 9, so B = 9.
- Tens: (A − 1) − 2 = 1 (the 1 is from borrowing). So A − 3 = 1, A = 4.
- Check: 48 − 29 = 19 ✓
Answer: A = 4, B = 9.
Example 6: Example 6: AAA pattern
Problem: Find A:
AA
+ AA
+ AA
———
CAB
Solution:
- 3 × AA = 3 × (11A) = 33A = CAB
- CAB is a 3-digit number, so 100 ≤ 33A ≤ 999.
- 33A ≥ 100 → A ≥ 4 (since 33 × 3 = 99, not 3-digit).
- Try A = 4: 33 × 4 = 132. C=1, A=4, B=2. But A should appear in result too: CAB = 142? Check: units digit should be A=4 but 132 has units 2. Hmm, CAB = C, A, B = 1, 3, 2. Wait — the result digits are C, A, B.
- Result = 132: C=1, second digit=3 (should be A), B=2. But A=4≠3. Not matching.
- Try A=8: 33×8=264. C=2, second digit should be A=8. But second digit is 6≠8. No.
- Try A=7: 33×7=231. C=2, second=3≠7. No.
- Try A=5: 33×5=165. C=1, second=6≠5. No.
- Try A=6: 33×6=198. C=1, second=9≠6. No.
- Try A=9: 33×9=297. C=2, second=9=A ✓, B=7.
- Check: 99+99+99=297. C=2, A=9, B=7. ✓
Answer: A = 9, B = 7, C = 2.
Example 7: Example 7: Two-letter addition
Problem: Find A:
8A
+ 8A
———
17A
Solution:
- 2 × (80 + A) = 170 + A
- 160 + 2A = 170 + A
- A = 10. But A must be 0-9!
- Wait — with carry: units: A + A. If 2A ≥ 10, there is carry.
- Units: 2A gives units digit A. So 2A ends in A.
- If A = 0: 2(0) = 0, units = 0 = A ✓. Tens: 8+8 = 16. Write 6 carry 1. But result tens = 7. 6≠7. No.
- 2A ends in A → A(2−1) = multiple of 10 → A = 0 (no carry) or from carry case: 2A−10 = A → A = 10 (impossible).
- Hmm. With carry from units: 2A = 10 + A → A = 10. Impossible.
- Without carry: 2A = A → A = 0. Tens: 8 + 8 = 16, write 6 carry 1. Hundreds = 1. Result = 160, but 17A = 170. 160≠170.
This puzzle has no solution — showing that not every letter puzzle is solvable.
Answer: No valid digit A satisfies this puzzle.
Example 8: Example 8: NCERT standard problem
Problem: Find the values of A and B if
A1
+ 1B
———
B0
Solution:
- Units: 1 + B. If 1+B gives units digit 0, then 1+B = 10, so B = 9.
- Carry = 1. Tens: A + 1 + 1 = B. So A + 2 = 9. A = 7.
- Check: 71 + 19 = 90 ✓. B0 = 90 ✓.
Answer: A = 7, B = 9.
Example 9: Example 9: Three-digit addition
Problem: Find A, B, C:
ABC
+ ABC
———
1A3A
Solution:
- 2 × ABC = 1A3A (a 4-digit number)
- ABC is a 3-digit number. 2 × ABC ranges from 200 to 1998.
- Result is 4-digit starting with 1, so 1000 ≤ 2×ABC ≤ 1999.
- Result = 1000 + 100A + 30 + A = 1030 + 101A.
- 2 × (100A + 10B + C) = 1030 + 101A
- 200A + 20B + 2C = 1030 + 101A
- 99A + 20B + 2C = 1030
- Try A = 9: 891 + 20B + 2C = 1030 → 20B + 2C = 139 → not possible (max = 20(9)+2(9)=198 but must be odd? 2C is even, 139 is odd). No.
- Try A = 8: 792 + 20B + 2C = 1030 → 20B + 2C = 238 → 10B + C = 119. Max = 99. No.
- Hmm, let me reconsider the result format: 1A3A means digits 1, A, 3, A.
- For A=5: 99(5)+20B+2C=1030 → 495+20B+2C=1030 → 20B+2C=535. Max=198. No.
- Actually, maybe the result means 1000+A×110+30+A = too simple. Let me try directly.
- 2 × 578 = 1156. Is this of form 1A3A? 1-1-5-6. No.
- Result 1A3A: try A=0: 1030. ABC=515. 2×515=1030. ✓ Digits: A=0 impossible (leading digit of ABC). Wait, A is hundreds digit of ABC=505? No: ABC with A=0 is 0BC, not 3-digit.
Let me try: A=5, result=1535. ABC=1535÷2=767.5. Not integer. A=6, result=1636. 1636÷2=818. ABC=818, A=8? But we said A=6. Contradiction.
This puzzle requires A in ABC to match A in result. If ABC=507, 2×507=1014=1A3A? 1-0-1-4. No.
Answer: This is a challenging puzzle that requires careful systematic checking of all values. In exam settings, simpler versions are given.
Example 10: Example 10: Product with repeated digit
Problem: Find B if B × 3 = B (units digit). What are the possible values of B?
Solution:
- We need 3B to end in B.
- B = 0: 3 × 0 = 0. Units digit = 0 = B. ✓
- B = 5: 3 × 5 = 15. Units digit = 5 = B. ✓
- Check others: B=1: 3, B=2: 6, B=3: 9, B=4: 12(2), B=6: 18(8), B=7: 21(1), B=8: 24(4), B=9: 27(7). None match.
Answer: B = 0 or B = 5.
Real-World Applications
Real-world applications:
- Logical reasoning: These puzzles build deductive thinking and systematic problem-solving skills.
- Competitive exams: Cryptarithmetic appears in Math Olympiads, NTSE, and other competitive tests.
- Computer science: Constraint-satisfaction problems in AI use similar logic to solve cryptarithmetic.
- Number theory: Understanding carries, place values, and modular arithmetic is reinforced through these puzzles.
Key Points to Remember
- Each letter represents exactly one digit (0-9).
- Different letters must represent different digits.
- The leading digit of any number cannot be 0.
- In addition, the maximum carry from any column is 1.
- Start solving from the column with the fewest unknowns.
- Use trial and elimination systematically.
- Always verify your answer by substituting back.
- Not all puzzles have solutions — this is a valid discovery.
Practice Problems
- Find A: A5 + 5A = 99.
- Find A and B: AB + BA = 132.
- Find A: 3A × A = 9A.
- Find A and B: A1 + 1B = B0.
- Find A: AA + AA + AA = 1AA. (Hint: 3 × 11A = 100 + 10A + A)
- Find A and B: AB − BA = 27.
Frequently Asked Questions
Q1. What are letters-for-digits puzzles?
These are puzzles where digits in arithmetic operations are replaced by letters. You find which digit each letter represents so the arithmetic is correct.
Q2. Can two different letters represent the same digit?
No. Each different letter must represent a different digit.
Q3. Can a number start with 0?
No. The leading (leftmost) digit of any number cannot be 0. So if A is the leading digit, A ≠ 0.
Q4. How do you solve these puzzles?
Use logic: start with the column that has the fewest unknowns (often the units column), consider possible carries, use elimination, and verify your answer.
Q5. What is the maximum carry in addition?
The maximum carry from adding a single column is 1 (since 9 + 9 + 1 = 19, and the carry is 1).
Q6. Do all puzzles have a unique solution?
Not necessarily. Some puzzles have multiple solutions, and some may have no solution at all.
Q7. What is cryptarithmetic?
Cryptarithmetic is another name for letters-for-digits puzzles. A famous example is SEND + MORE = MONEY.
Q8. How are these related to the general form of numbers?
Each number in the puzzle can be written in general form (e.g., AB = 10A + B), converting the puzzle into algebraic equations.
