Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 5: Number Play (Ganita Prakash)
Authentic handwritten solutions in Topper Notebook style. Features complete divisibility proofs, digit manipulations, cryptarithm solutions, Venn diagram set relations, and parity analyses for all 16 exercises from Pages 33 to 37 of Ganita Prakash Part I.
Quick Chapter Navigation & Blueprint
- Part A: Divisibility Tests & Digit Puzzles (Questions 4 to 8) — Multiples of 18 and 44, finding missing digits (a, b) and (p, q), consecutive multiples, and multiples of 36.
- Part B: Sequences, Reversals & Conjectures (Questions 9 to 12) — 5 consecutive even numbers in terms of p, 6-digit reversible divisibility (15 & 6), Deepak’s doubling claim, and Always/Sometimes/Never True statements.
- Part C: Modulo Arithmetic & Factorials (Questions 13 & 14) — Sum of 3 numbers divisible by 3 (remainder cases) and products of 2, 3, 4, and 5 consecutive integers.
- Part D: Cryptarithms & Venn Diagrams (Questions 15 & 16) — Solving EF × E = GGG and WOW × 5 = MEOW, and concentric Venn diagram hierarchy for multiples of 4, 8, and 32.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 5 (hegp105.pdf)
Part A: Divisibility Rules & Digit Investigations (Pages 33–34)
Question 5 (Page 33):
If 48a23b is a multiple of 18, list all possible pairs of values for digits a and b.
If 48a23b is a multiple of 18, list all possible pairs of values for digits a and b.
Topper’s Two-Condition Divisibility Breakdown:
- A number is divisible by 18 if and only if it is divisible by both 2 and 9 (since 18 = 2 × 9 and gcd(2, 9) = 1).
- Condition 1 (Divisibility by 2): The units digit b must be an even digit:
b ∈ {0, 2, 4, 6, 8}. - Condition 2 (Divisibility by 9): The sum of all digits must be a multiple of 9:
Sum = 4 + 8 + a + 2 + 3 + b = 17 + a + b. - Testing each possible value of b (where 0 ≤ a ≤ 9):
• If b = 0: Sum = 17 + a → 17 + a = 18 → a = 1. (Pair: 1, 0)
• If b = 2: Sum = 19 + a → 19 + a = 27 → a = 8. (Pair: 8, 2)
• If b = 4: Sum = 21 + a → 21 + a = 27 → a = 6. (Pair: 6, 4)
• If b = 6: Sum = 23 + a → 23 + a = 27 → a = 4. (Pair: 4, 6)
• If b = 8: Sum = 25 + a → 25 + a = 27 → a = 2. (Pair: 2, 8)
All Possible Pairs (a, b):
(1, 0), (8, 2), (6, 4), (4, 6), and (2, 8).
(1, 0), (8, 2), (6, 4), (4, 6), and (2, 8).
Question 6 (Page 33–34):
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Divisibility by 4 & 11 Combined:
- A number is divisible by 44 if it is divisible by both 4 and 11.
- Condition 1 (Divisibility by 4): The last two digits ‘q8’ must form a multiple of 4:
Possible 2-digit combinations ending in 8: 08, 28, 48, 68, 88 → q ∈ {0, 2, 4, 6, 8}. - Condition 2 (Divisibility by 11): The difference between the sum of odd-place digits and even-place digits must be 0 or a multiple of 11:
• Sum of odd-place digits (1st, 3rd, 5th from right): 8 + 7 + 3 = 18.
• Sum of even-place digits (2nd, 4th from right): p + q.
• Difference = 18 − (p + q). - Since 0 ≤ p, q ≤ 9 → 0 ≤ p + q ≤ 18:
Case 1: 18 − (p + q) = 11 → p + q = 7.
• If q = 0 → p = 7 → Pair: (7, 0)
• If q = 2 → p = 5 → Pair: (5, 2)
• If q = 4 → p = 3 → Pair: (3, 4)
• If q = 6 → p = 1 → Pair: (1, 6)
• If q = 8 → p = −1 (not a valid digit). - Case 2: 18 − (p + q) = 0 → p + q = 18 → only possible if p = 9, q = 9, but q must be even! So no solutions here.
All Possible Pairs (p, q):
(7, 0), (5, 2), (3, 4), and (1, 6).
(7, 0), (5, 2), (3, 4), and (1, 6).
Question 7 (Page 34):
Find three consecutive numbers such that the first number is a multiple of 2, the second is a multiple of 3, and the third is a multiple of 4. Are there more such numbers? How often do they occur?
Find three consecutive numbers such that the first number is a multiple of 2, the second is a multiple of 3, and the third is a multiple of 4. Are there more such numbers? How often do they occur?
Topper’s LCM Periodicity Analysis:
- Let the three consecutive numbers be n, n + 1, and n + 2.
- Conditions:
• n is a multiple of 2 (i.e., n is even).
• n + 1 is a multiple of 3.
• n + 2 is a multiple of 4 → implies n leaves remainder 2 when divided by 4. - Testing the smallest positive integers:
For n = 2:
• 1st number = 2 (multiple of 2) ✓
• 2nd number = 3 (multiple of 3) ✓
• 3rd number = 4 (multiple of 4) ✓ - How often do they occur?
LCM(2, 3, 4) = 12. Since the conditions are periodic with period 12, this pattern repeats every 12 integers:
• 2, 3, 4
• 2 + 12 = 14, 15, 16 (14 is multiple of 2, 15 is mult of 3, 16 is mult of 4) ✓
• 14 + 12 = 26, 27, 28 ✓
• 26 + 12 = 38, 39, 40 ✓
Answers:
Smallest set: 2, 3, 4.
Yes, there are infinitely many such numbers; they occur once every 12 numbers (e.g., 14, 15, 16; 26, 27, 28; etc.).
Smallest set: 2, 3, 4.
Yes, there are infinitely many such numbers; they occur once every 12 numbers (e.g., 14, 15, 16; 26, 27, 28; etc.).
Question 8 (Page 34):
Write five multiples of 36 between 45,000 and 47,000.
Write five multiples of 36 between 45,000 and 47,000.
Step-by-Step Generation:
- Dividing 45,000 by 36:
45000 ÷ 36 = 1250 (exact integer!). - So 45,000 itself is 36 × 1250.
- The subsequent multiples strictly between 45,000 and 47,000 are obtained by adding 36 repeatedly:
• 1st multiple: 45000 + 36 = 45,036
• 2nd multiple: 45036 + 36 = 45,072
• 3rd multiple: 45072 + 36 = 45,108
• 4th multiple: 45108 + 36 = 45,144
• 5th multiple: 45144 + 36 = 45,180
Five Multiples of 36:
45036, 45072, 45108, 45144, and 45180.
45036, 45072, 45108, 45144, and 45180.
Part B: Sequences, Reversals & Logic (Pages 34–35)
Question 9 (Page 34):
The middle number in a sequence of 5 consecutive even numbers is 5p. Express the other four numbers in the sequence in terms of p.
The middle number in a sequence of 5 consecutive even numbers is 5p. Express the other four numbers in the sequence in terms of p.
Even Number Gap Analysis:
- Consecutive even numbers always differ by 2.
- Let the 5 consecutive even numbers arranged in ascending order be:
T1, T2, T3, T4, T5. - Given: The middle number T3 = 5p.
- Therefore:
• T1 = 5p − 4
• T2 = 5p − 2
• T3 = 5p
• T4 = 5p + 2
• T5 = 5p + 4
The other four numbers are: 5p − 4, 5p − 2, 5p + 2, and 5p + 4.
Question 10 (Page 34–35):
Write a 6-digit number that is divisible by 15, such that when its digits are reversed, it is divisible by 6.
Write a 6-digit number that is divisible by 15, such that when its digits are reversed, it is divisible by 6.
Topper’s Deduction Method:
- Let the 6-digit number be N = abcdef.
- Condition 1: Divisible by 15:
• Must end in 0 or 5 (i.e., f = 0 or f = 5).
• If f = 0, the reversed number starts with 0 (0edcba), which is only a 5-digit number! Hence, f = 5. - Condition 2: Reversed number 5edcba is divisible by 6:
• Must be divisible by 2 → its last digit a must be even: a ∈ {2, 4, 6, 8} (since a ≠ 0).
• Must be divisible by 3 → sum of digits (a + b + c + d + e + 5) must be a multiple of 3. - Let a = 2, b = 0, c = 0, d = 0, f = 5:
Sum = 2 + 0 + 0 + 0 + e + 5 = 7 + e.
For 7 + e to be divisible by 3, choose e = 2 (sum = 9).
The original number is: 200025.
• Check 1: 200025 ends in 5, sum of digits = 9 (divisible by 15) ✓
• Check 2: Reversed is 520002, ends in 2, sum = 9 (divisible by 6) ✓
Example 6-digit Number: 200025 (other valid examples include 200055, 200085, 400035, 600015).
Question 11 (Page 35):
Deepak claims: “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled.” Examine if his conjecture is true.
Deepak claims: “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled.” Examine if his conjecture is true.
Algebraic Proof of Falsehood:
- Let n be any multiple of 11. By definition, n = 11 × k, for some integer k.
- When doubled, the new number is:
2n = 2 × (11 × k) = 11 × (2k). - Since k is an integer, 2k is also an integer.
- Therefore, 2n is unconditionally a multiple of 11 for EVERY multiple of 11.
Conclusion: Deepak’s conjecture is FALSE. Doubling any multiple of 11 will always result in a multiple of 11.
Question 12 (Page 35 – True/False Logic):
Determine whether the following are Always True, Sometimes True, or Never True:
(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 abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.
Determine whether the following are Always True, Sometimes True, or Never True:
(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 abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.
Topper’s Rigorous Analysis:
- (i) Always True: Multiple of 6 = 6x = 3(2x). Multiple of 3 = 3y. Product = (6x)(3y) = 18xy = 9(2xy), which is always a multiple of 9.
- (ii) Always True: Three consecutive even numbers are 2k − 2, 2k, 2k + 2. Sum = 6k, which is always divisible by 6.
- (iii) Always True: Divisibility by 6 requires divisibility by 2 and 3. In both abcdef and badcef, the units digit is f (same parity), and the set of digits is identical (same sum of digits). Hence, both are always simultaneously divisible by 6.
- (iv) Never True: Expanding: 8(7b − 3) − 4(11b + 1) = 56b − 24 − 44b − 4 = 12b − 28 = 12(b − 2) − 4. Dividing by 12 leaves a remainder of 8 for every integer b. Thus, it can NEVER be a multiple of 12.
Answers:
(i) Always True | (ii) Always True | (iii) Always True | (iv) Never True.
(i) Always True | (ii) Always True | (iii) Always True | (iv) Never True.
Part C: Modulo Classes, Cryptarithms & Venn Diagrams (Pages 35–37)
Question 13 (Page 35–36):
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Topper’s Remainder Modulo-3 Analysis:
- When any integer is divided by 3, the only possible remainders are 0, 1, or 2.
- Let the remainders of the three chosen numbers be r1, r2, and r3.
- The sum n1 + n2 + n3 is divisible by 3 if and only if r1 + r2 + r3 is a multiple of 3 (0, 3, or 6).
- Case 1 (All remainders identical):
• All remainders 0 → 0 + 0 + 0 = 0 (divisible by 3)
• All remainders 1 → 1 + 1 + 1 = 3 (divisible by 3)
• All remainders 2 → 2 + 2 + 2 = 6 (divisible by 3) - Case 2 (All three remainders distinct):
• One of each: 0 + 1 + 2 = 3 (divisible by 3) - All other combinations (e.g. two 1s and one 0 → sum = 2) fail.
Generalisation: The sum of 3 numbers is divisible by 3 if and only if:
1. All three numbers leave the SAME remainder when divided by 3, OR
2. All three numbers leave DIFFERENT remainders (one each of remainder 0, 1, and 2).
1. All three numbers leave the SAME remainder when divided by 3, OR
2. All three numbers leave DIFFERENT remainders (one each of remainder 0, 1, and 2).
Question 14 (Page 36–37):
Is the product of two consecutive integers always a multiple of 2? What about three consecutive integers (multiple of 6)? Four consecutive integers (multiple of 24)? Five consecutive integers (multiple of 120)? Explain.
Is the product of two consecutive integers always a multiple of 2? What about three consecutive integers (multiple of 6)? Four consecutive integers (multiple of 24)? Five consecutive integers (multiple of 120)? Explain.
Consecutive Integer Factorial Divisibility:
- Two Consecutive Integers n(n + 1): One of any two consecutive numbers is always even. Thus, the product is always a multiple of 2 (= 2!).
- Three Consecutive Integers n(n + 1)(n + 2): Among any three consecutive numbers, at least one is a multiple of 2 and exactly one is a multiple of 3. Since gcd(2, 3) = 1, the product is always a multiple of 2 × 3 = 6 (= 3!).
- Four Consecutive Integers: Contains at least two even numbers (one multiple of 2, one multiple of 4) and at least one multiple of 3. The product is always a multiple of 2 × 4 × 3 = 24 (= 4!).
- Five Consecutive Integers: Contains factors contributing multiples of 2, 3, 4, and 5. The product is always a multiple of 24 × 5 = 120 (= 5!).
General Theorem: The product of any k consecutive integers is always divisible by k! (k factorial).
Question 15 (Page 37 – Cryptarithms):
Solve the cryptarithms where each distinct letter represents a unique single digit (0–9):
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Solve the cryptarithms where each distinct letter represents a unique single digit (0–9):
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Cryptarithm Deduction Steps:
- (i) Solving EF × E = GGG:
GGG represents a 3-digit repdigit: 111, 222, 333, 444, 555, 666, 777, 888, 999.
Notice that GGG = G × 111 = G × 3 × 37.
Since 37 is prime, one of the factors must be 37!
Hence, 2-digit number EF = 37 → E = 3, F = 7.
Then EF × E = 37 × 3 = 111 → G = 1. - (ii) Solving WOW × 5 = MEOW:
• A 3-digit number multiplied by 5 gives a 4-digit number ending in W.
• The product of any integer and 5 must end in 0 or 5. Since W is the leading digit of WOW, W ≠ 0. Therefore, W = 5!
• Now equation is: 5O5 × 5 = ME5O.
• 5 × 5 = 25 → last digit is 5 (matches W = 5), carry-over is 2.
• Tens digit: 5 × O + 2 ends in O. Testing digits:
If O = 7: 5 × 7 + 2 = 37 → ends in 7! (So O = 7, carry-over = 3).
• Hundreds and thousands digits: 5 × 5 + 3 = 28 → ME = 28 → M = 2, E = 8.
• Check: 575 × 5 = 2875 ✓ (Matches MEOW: M=2, E=8, O=7, W=5).
Cryptarithm Solutions:
(i) E = 3, F = 7, G = 1 (37 × 3 = 111)
(ii) W = 5, O = 7, M = 2, E = 8 (575 × 5 = 2875)
(i) E = 3, F = 7, G = 1 (37 × 3 = 111)
(ii) W = 5, O = 7, M = 2, E = 8 (575 × 5 = 2875)
Question 16 (Page 37 – Multiples Venn Diagram):
Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Subset Inclusion Analysis:
- Every multiple of 32 is a multiple of 8 (since 32 = 8 × 4). Thus:
{Multiples of 32} ⊆ {Multiples of 8}. - Every multiple of 8 is a multiple of 4 (since 8 = 4 × 2). Thus:
{Multiples of 8} ⊆ {Multiples of 4}. - Combining these inclusions gives a nested chain of subsets:
{Multiples of 32} ⊆ {Multiples of 8} ⊆ {Multiples of 4}. - In a Venn diagram, this represents three concentric circles, with Multiples of 32 in the innermost circle, surrounded by Multiples of 8, and Multiples of 4 as the outermost set.
Correct Option: Figure (iv) — Concentric circles representing nested subsets.