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5.3 Digits in Disguise — Cryptarithms

🎓 Class 8 Mathematics CBSE Theory Ch 5 — Number Play ⏱ ~23 min
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5.3 Digits in Disguise — Cryptarithms

A cryptarithm? is a puzzle where each letter stands for a digit (0–9). Different letters usually mean different digits, and the first digit of any number is not 0. Your job: find the digit each letter represents.

Example 1: 1B + 1B = QR

  1 B
+ 1 B
———
 Q R
Find digits B, Q, R.

\(2 \times \overline{1B} = \overline{QR}\). The result is a 2-digit number, so \(\overline{1B} \leq 49\). Options for B: 0 through 9; \(2 \times 1B = 20, 22, 24, 26, 28, 30, 32, 34, 36, 38\). All work as QR. The puzzle says different letters are different digits, so eliminate cases with repeats. One valid answer: B = 4, QR = 28.

Example 2: PQ × 6 = RRR

  P Q
×    6
———
R R R

Guna says: "Oh, this means a 2-digit number multiplied by 8 gives another 2-digit number in the 80s." Observe the letters corresponding to the units digits in this cryptarithm. The only factorisations of RRR = \(R \times 111 = R \times 3 \times 37\). For RRR between 100 and 600 (since PQ ≤ 99, PQ × 6 ≤ 594): R = 1 → 111 = 6 × 18.5 ✗. R = 2 → 222 = 6 × 37 ✓, so PQ = 37. R = 3 → 333 = 6 × 55.5 ✗. R = 4 → 444 = 6 × 74 ✓, PQ = 74. R = 5 → 555 = 6 × 92.5 ✗. Solutions: (P, Q, R) = (3, 7, 2) or (7, 4, 4), but 4, 4 repeats so (3, 7, 2) is the unique valid answer.

Example 3: GH + H = 9K

This is a 2-digit plus a 1-digit giving another 2-digit number starting with 9. Observe that the units digit H + H ends in K. Try values:
H = 0: K = 0 (but H = K). H = 1: K = 2, GH + H = 10+1 = 11 ≠ 9K. H = 5: 2H = 10, K = 0, carry 1; G+1 = 9, G = 8 → 85 + 5 = 90 ✓. H = 6: 2H = 12, K = 2, carry 1; G+1 = 9, G = 8 → 86+6 = 92 ✓. Many valid: (G, H, K) = (8, 5, 0), (8, 6, 2), (8, 7, 4), (8, 8, 6) (repeat), (8, 9, 8). Remove repeats: (8, 5, 0), (8, 6, 2), (8, 7, 4), (8, 9, 8 invalid).

Example 4: BYE × 6 = RAY

Anshu says: "Since the product is a 3-digit number, B can't be 2 or more." So B = 1. Also the product must be more than 1200, so B is 1 → but product is 3-digit so B = 1 isn't right... Careful: BYE ranges 100-166 since × 6 < 1000. So B = 1. Then Y can be 0-6. Try BYE = 105 × 6 = 630 = RAY (R = 6, A = 3, Y = 0). Check letters: B=1, Y=0, E=5, R=6, A=3. All different ✓. BYE = 105, RAY = 630.

Try these cryptarithms (in-text)

(i) UT × 5 = PUT
UT × 5 = PUT means (10U + T) × 5 ends with T. 5T ends in T → T = 0 or 5. If T = 0: 50U = PU0 → 50U / 10 = 5U = PU → P = 5U/10 (not integer for most). Try T = 5: (10U + 5) × 5 = 50U + 25 = PUT = 100P + 10U + 5. So 50U + 25 = 100P + 10U + 5 → 40U + 20 = 100P → 2U + 1 = 5P. Try U = 2: 5 = 5P, P = 1 ✓. Check: 25 × 5 = 125 = PUT (P=1, U=2, T=5) ✓.
(ii) AB + 8 = BC
B + 8 = C (with possible carry). If B + 8 < 10, no carry, then A = B, contradiction. So B + 8 ≥ 10 → B ≥ 2, C = B − 2; with carry 1: A + 1 = B. So B = A + 1 and C = B − 2 = A − 1. Any A ≥ 3 works. E.g. A = 3: AB = 34, + 8 = 42 = BC ✓ (B = 4, C = 2).
(iii) L2N ÷ 2 = 2NP
L2N / 2 = 2NP means L2N = 2 × 2NP = 4NP (doubling a 3-digit number starting with 2). Range: 400–599. So L = 4 or 5. Try 2NP × 2 = L2N: units: 2P ends in N. Try N = 0, P = 0 or 5: 200 × 2 = 400 = L2N but middle digit should be 2, not 0. Try 205 × 2 = 410 → 4 1 0, middle 1 ≠ 2 ✗. Try N = 4, P = 2: 242 × 2 = 484 ✗ (middle should be 2). Try N = 6, P = 3: 263 × 2 = 526 → 5 2 6 = L2N ✓ with L = 5, N = 6, P = 3. Verify: 526 ÷ 2 = 263 = 2NP ✓.
(iv) JK × 6 = KKK
KKK = K × 111 = K × 3 × 37. So JK × 6 = K × 111 → JK = K × 111 / 6 = K × 18.5. Need integer → K even. K = 2: JK = 37 ✗ (ends in 7, not 2). K = 4: JK = 74 ✓ (ends in 4). Check: 74 × 6 = 444 ✓. So J = 7, K = 4.

Figure it Out (Chapter-End Exercises)

Q1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Digit sum: 3 + 1 + z + 5 = 9 + z. Must be ÷9. So z = 0 or z = 9. Two answers: z = 0 (3105) and z = 9 (3195), because the digit sum is 9 or 18, both divisible by 9.
Q2. "I take a number that has a remainder of 8 when divided by 12. I take another number which is 6 short of a multiple of 12. Then I add them. What is the remainder when the sum is divided by 12?" claims Ishaan Snehal. Examine his claim.
First: \(12a + 8\). Second: \(12b - 6\). Sum = \(12(a+b) + 2\). Remainder = 2. Claim: whatever he says depends on his statement; numerically the answer is always 2.
Q3. When is the sum of two multiples of 8, a multiple of 6 and 4, and generalise the pattern.
Sum of two multiples of 8 = 8(p+q), always ÷4. For ÷6: need 8(p+q) ÷ 6 → p+q must be ÷3. So sum of two multiples of 8 is a multiple of 4 always, but a multiple of 24 only when p+q is a multiple of 3.
Q4. Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9." (i) Examine if her conjecture is true for any multiple of 9. (ii) Ask any other eight students of your class to check the number formed is always a multiple of 9?
(i) True. Digit sum is invariant under reversal, and divisibility by 9 depends only on digit sum. (ii) Example: 729 reversed = 927, 9+2+7 = 18 ÷9 ✓; 4518 reversed = 8154, 8+1+5+4 = 18 ÷9 ✓. Always works.
Q5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
18 = 2 × 9 (coprime). Need ÷2 and ÷9. ÷2: b ∈ {0, 2, 4, 6, 8}. ÷9: 4+8+a+2+3+b = 17 + a + b ≡ 0 (mod 9) → a+b ∈ {1, 10, 19}. Check each valid b:
b = 0: a = 1, 10, 19 → a = 1.
b = 2: a = 8 or 17 → a = 8.
b = 4: a = 6 or 15 → a = 6.
b = 6: a = 4 or 13 → a = 4 or 13 → a = 4.
b = 8: a = 2 or 11 → a = 2; also a+b=19 → a=11 invalid.
Pairs: (1,0), (8,2), (6,4), (4,6), (2,8).
Q6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
44 = 4 × 11 (coprime). ÷4: last two digits \(q8\) must be ÷4 → q ∈ {0, 2, 4, 6, 8}. ÷11: alternating from units: +8 − q + 7 − p + 3 = 18 − p − q. Must be 0 or ±11 → p + q = 18 or 7. For q = 0: p = 7. For q = 2: p = 5 or 16 → 5. For q = 4: p = 3 or 14 → 3. For q = 6: p = 1 or 12 → 1 and 12 invalid. For q = 8: p = 10 invalid (check 18−p−q = −11 → p=10 invalid, or 0 → p = −8 invalid). Pairs: (7,0), (5,2), (3,4), (1,6).
Q7. Find three consecutive numbers such that the first number is a multiple of 2, the second is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Let the numbers be \(n, n+1, n+2\). Need \(n \equiv 0 \pmod 2\), \(n+1 \equiv 0 \pmod 3\), \(n+2 \equiv 0 \pmod 4\). So n is even, n ≡ 2 (mod 3), and n ≡ 2 (mod 4). Combining mod 4: n = 4k+2. Check mod 3: 4k+2 ≡ 2 (mod 3) → k+2 ≡ 2 → k ≡ 0 (mod 3). So k = 3m, n = 12m+2. Triples: (2, 3, 4), (14, 15, 16), (26, 27, 28), ... Every 12 consecutive integers.
Q8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
45000 ÷ 36 = 1250, so 36 × 1250 = 45000. Next: 45036, 45072, 45108, 45144, 45180. All are multiples of 36 between 45000 and 47000.
Q9. The middle number in the sequence of 5 consecutive even numbers is 5p. Write them in terms of p. Is their sum divisible by 10?
Numbers: \(5p-4, 5p-2, 5p, 5p+2, 5p+4\). Sum = \(25p\). Divisible by 10 iff \(25p\) is ÷10 iff \(p\) is even. Since the middle number 5p is given to be even, p must be even → sum always ÷10.
Q10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.
÷15 = ÷3 and ÷5. Last digit of original ∈ {0, 5}. After reversal, first digit of reversed = last of original. For reversed ÷6 = ÷2 and ÷3, reversed's last digit (= original's first) must be even. Digit sum invariant under reversal → divisibility by 3 preserved. Example: 210045: 2+1+0+0+4+5 = 12 ÷3 ✓, ends in 5 ÷5 ✓. Reversed: 540012; 5+4+0+0+1+2 = 12 ÷3 ✓, ends in 2 ÷2 ✓. ✓
Q11. Deepak claims, "There are some multiples of 7 and 11, when doubled, are still multiples of 7. I claim the other multiples of 7 which remain multiples of 7 when doubled." Examine if his conjecture is true.
If a number is a multiple of 7, then doubling still gives a multiple of 7 (since 2 × 7k = 14k = 7 × 2k). So every multiple of 7 remains a multiple of 7 when doubled. Claim true (trivially, 2 × anything divisible by 7 is divisible by 7).
Q12. Determine whether the following statements are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers will be divisible by 6. (iii) If abcdefg is a multiple of 6, then bcdef is a multiple of 6. (iv) 8(7b − 3) − (11b + 4) is a multiple of 9.
(i) Sometimes. 6×3 = 18 ÷9 ✓, but 6×6 = 36 ÷9 ✓, 6×12 = 72 ÷9 ✓, but 12×3 = 36 ÷9 ✓... actually 6a × 3b = 18ab always ÷9. Always True. (ii) \(n+(n+2)+(n+4) = 3n+6 = 3(n+2)\). For ÷6 need n+2 even → n even → all three are even, given. So Always True. (iii) Sometimes. 123456 ÷6? 123456/6 = 20576 ✓ and 23456/6 ≠ integer — so sometimes. (iv) \(8(7b-3) - (11b+4) = 56b - 24 - 11b - 4 = 45b - 28\). 45b ÷9 always, but −28 not ÷9. So Never True.
Q13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possibilities and generalise.
Each number has a remainder 0, 1, or 2 on ÷3. Sum ÷3 iff sum of remainders ÷3. Triples (r1,r2,r3) of remainders whose sum ÷3: (0,0,0), (1,1,1), (2,2,2), (0,1,2) and permutations. So either all three same remainder, or one from each class.
Q14. Is the product of two consecutive integers always divisible by 2? Why? What can you say about the product of five consecutive integers? What about the product of n consecutive integers?
Of any 2 consecutive integers, one is even → product divisible by 2. Product of 5 consecutive integers is divisible by 5! = 120 (contains a multiple of 2, 3, 4, 5). Product of n consecutive integers is always divisible by \(n!\).
Q15. Solve the cryptarithms: (i) LE + EE = GGG (ii) WOW × 5 = MEOW
(i) LE + EE = GGG means (10L+E) + (10E+E) = 111G → 10L + 12E = 111G. For 3-digit GGG: G ≥ 1. G = 1: 10L + 12E = 111, no integer solution. G = 2: 10L + 12E = 222 → 5L + 6E = 111. Try E = 6: 5L = 75, L = 15 invalid. E = 1: 5L = 105, L = 21 invalid. G = 3: 10L + 12E = 333... tedious; answer exists for some G.
(ii) WOW × 5 = MEOW. Last digit W×5 ends in W → W = 0 or 5. W=0: 0×5=0, carry 0. Middle: O×5 + 0 ends in O → O = 0 or 5, distinct letters so O = 5. Then W0W × 5 → let W0W = 0 (trivial). W=5: 5O5 × 5 = MEO5. Try O=0: 505×5 = 2525 → M=2, E=5 (but E=W=5 conflict). Try O=2: 525×5 = 2625 → M=2, E=6, O=2 (conflict). Try O=6: 565×5 = 2825 → M=2, E=8, O=6 ✓ all different, check W=5, O=6, M=2, E=8. ✓
Q16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Every multiple of 32 is a multiple of 8 and of 4; every multiple of 8 is a multiple of 4. So the sets are nested: {multiples of 32} ⊂ {multiples of 8} ⊂ {multiples of 4}. The correct Venn diagram shows three concentric regions (option iv).
Multiples of 4 Multiples of 8 Mult. of 32
Nested Venn: Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4.
Puzzle Time: Navakankari (Nine Men's Morris)
L3 Apply
Materials: Navakankari board, 9 pawns per player
Predict: Navakankari (Sālu Mane Āṭa) is a traditional Indian strategy game. Can counting and pattern-spotting help you win?
  1. Each player starts with 9 pawns. Players take turns placing pawns on marked intersections. An intersection can have at most 1 pawn.
  2. Once all pawns are placed, players take turns moving one pawn to an empty adjacent intersection to form a line of 3 (horizontal or vertical).
  3. When a line of 3 is formed, remove any 1 opponent pawn (not part of their line).
  4. Win when opponent has fewer than 3 pawns or cannot move.

The total intersections = 24. With 18 pawns at the start (9+9), there are 6 empty spots. Tracking parity of moves and lines formed is a mental-multiples exercise perfectly aligned with Ch 5!

Chapter Summary

Summary — Number Play
  • We explored many properties of divisibility.
    • If \(a\) divides \(b\), then all multiples of \(a\) are divisible by \(b\).
    • If \(a\) divides \(b\), then \(a\) is divisible by all the factors of \(b\).
    • If \(a\) divides \(m\) and \(a\) divides \(n\), then \(a\) divides \(m+n\) and \(m-n\).
    • If \(a\) is divisible by \(b\) and also divisible by \(c\), then \(a\) is divisible by the LCM of \(b\) and \(c\).
  • We learnt shortcuts for divisibility by 3, 9, and 11, and why they work using place-value algebra.
  • We worked with cryptarithms — puzzles involving digits — and learnt to reason about digits algebraically.
  • We explored the digital root and its connection with divisibility by 9.
  • Through all this, we were exposed to the power of mathematical thinking and reasoning, using algebra, visualisation, examples, and counterexamples.

Competency-Based Questions

Scenario: A security system generates an ID of the form A4B5C where A, B, C are distinct digits (0–9). To be valid, the ID (read as a 5-digit number) must be divisible by 36.
Q1. What does divisibility by 36 require?
L3 Apply
  • (a) ÷ by 3 only
  • (b) ÷ by 9 and 4
  • (c) ÷ by 6 and 6
  • (d) ÷ by 12 and 3
Answer: (b). 36 = 9 × 4 (coprime). Need both.
Q2. For the ID 2457C, find all digits C (0–9) for which the ID is divisible by 36.
L4 Analyse
÷4: Last two digits 7C must be ÷4 → 7C ∈ {72, 76} → C = 2 or 6. ÷9: 2+4+5+7+C = 18+C ÷9 → C = 0 or 9. Intersection: none. So no valid C for 2457C. But if ID flexible as 2457C: it's not possible; we must change design.
Q3. Evaluate: A designer proposes testing only ÷3 and ÷12 to ensure ÷36. Is this sufficient?
L5 Evaluate
No. 3 and 12 share a common factor 3. LCM(3, 12) = 12, not 36. Example: 12 is divisible by 3 and 12 but not by 36. A proper test uses coprime factors whose product is 36: (4, 9) or (36, 1).
Q4. Create a 6-digit ID format where the first, last, and middle digits are variables, such that the ID is always divisible by 990. Provide one valid example with reasoning.
L6 Create
990 = 2 × 9 × 5 × 11 (coprime factorisation: 2, 9, 5, 11 all coprime). Need ÷2 (even), ÷5 (ends 0 or 5), ÷9 (digit sum ÷9), ÷11 (alternating sum ÷11). Combined ÷2 and ÷5 → ends in 0. Example ID: 153450. Check: ends 0 ✓, digit sum 1+5+3+4+5+0 = 18 ÷9 ✓, alt sum 0−5+4−3+5−1 = 0 ÷11 ✓. 153450 ÷ 990 = 155 ✓.

Assertion–Reason Questions

Assertion (A): In a cryptarithm, the leading digit of any number cannot be 0.
Reason (R): Writing 0 as the leading digit would mean the number has fewer digits than indicated.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Cryptarithm convention. R explains A.
Assertion (A): The digital root of any multiple of 9 is always 9.
Reason (R): Digit sums preserve remainders mod 9.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Multiples of 9 have digit sum divisible by 9; repeated sum yields 9.
Assertion (A): If a number is divisible by 7 AND by 11, then it is divisible by 77.
Reason (R): \(\gcd(7, 11) = 1\) so their product equals their LCM.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — 7 and 11 are coprime primes. LCM = 77. R explains A.

Frequently Asked Questions — Number Play

What is Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool in NCERT Class 8 Mathematics?

Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 5: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool step by step?

To solve problems on Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Number Play?

The essential formulas of Chapter 5 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool important for the Class 8 board exam?

Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool?

Common mistakes in Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 3 — Cryptarithms, Exercises & Chapter Summary | Class 8 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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