Class 8 Maths Ganita Prakash Chapter 5 Solutions (NCERT 2026–27) – Number Play

These Class 8 Maths Ganita Prakash Chapter 5 solutions cover Number Play from the new NCF-2023 textbook (Part I, 2026–27). Every “Figure it Out” exercise is solved step by step — parity, sums of consecutive numbers, divisibility shortcuts for 3, 9 and 11, digital roots and cryptarithms — so you can master each idea and revise the whole chapter quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 5 Exercises: 4 × Figure it Out Session: 2026–27
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Chapter 5 Overview

Chapter 5 of Ganita Prakash, Number Play, is an exploration of how numbers behave under the operations of addition, subtraction and multiplication. It begins with sums of consecutive numbers and the idea of parity (odd/even), then builds up powerful divisibility properties — if a number is divisible by another, its multiples and factors behave predictably, and the sum or difference of two multiples of a number is again a multiple. The chapter develops and proves the divisibility shortcuts for 3, 9 and 11 using place value and visualisation, introduces digital roots, and ends with cryptarithms (digit puzzles). The emphasis throughout is on reasoning — understanding “why” a rule works using algebra, examples and counterexamples. The Class 8 Maths Ganita Prakash Chapter 5 solutions below work through every “Figure it Out” question with verified, exam-ready steps.

Key Concepts & Definitions

Parity: whether a number is even (a multiple of 2) or odd. Negative even numbers (−2, −4, …) are also even.

Consecutive numbers: whole numbers that follow one another, e.g. n, n+1, n+2, …

Multiple & factor: if a = b × k, then a is a multiple of b, and b is a factor of a.

Remainder form: a number leaving remainder r on division by m is written as mk + r (e.g. 5k + 3 for remainder 3 mod 5).

Digital root: the single digit obtained by repeatedly adding the digits of a number; for a positive whole number, roots 1–8 equal the remainder on division by 9, while root 9 means remainder 0. The digital root of 0 is 0.

Cryptarithm: a puzzle where each letter stands for one digit, each digit by at most one letter, and a leading digit is never 0.

Important Facts & Divisibility Rules (Chapter 5)

Parity of sums/differences: odd ± odd = even, even ± even = even, odd ± even = odd. So a ± b always have the same parity.

If a divides M and a divides N, then a divides (M + N) and (M − N).

If A is divisible by k, then all multiples of A are divisible by k, and A is divisible by every factor of k.

If A is divisible by k and by m, then A is divisible by LCM(k, m).

Divisibility by 9 / 3: a number is divisible by 9 (or 3) if and only if the sum of its digits is divisible by 9 (or 3). For a positive number, digital root 9 means remainder 0; other digital roots equal the remainder.

Divisibility by 11: form the alternating sum of digits starting with a plus at the units place (+, −, +, −, … when moving left); the number is divisible by 11 if this result is 0 or a multiple of 11.

Figure it Out (Page 122) — Class 8 Maths Ganita Prakash Chapter 5 Solutions

1. The sum of four consecutive numbers is 34. What are these numbers?

SOLUTION Let the four consecutive numbers be n, n+1, n+2, n+3. Sum = 4n + 6 = 34 ⇒ 4n = 28 ⇒ n = 7. So the numbers are 7, 8, 9 and 10 (check: 7 + 8 + 9 + 10 = 34). ✓

2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

SOLUTION Since p is the greatest, each earlier number is 1 less than the previous one. The other four numbers are (p − 1), (p − 2), (p − 3) and (p − 4).

3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

SOLUTION(i) Sometimes true. Two even numbers are 2m and 2n, with sum 2(m + n). This is divisible by 3 exactly when m + n is divisible by 3. Example: 2 + 4 = 6. Non-example: 2 + 6 = 8.(ii) Sometimes true. A number 9k with odd k is divisible by 9 but not 18, so 27 is a counterexample. A number such as 30 is divisible by neither, so the statement holds for it. If a multiple of 9 is also even, it is a multiple of 18.(iii) Sometimes true. Write the numbers as 6a + r and 6b + s, where r and s lie from 1 to 5. Their sum is 6(a + b) + r + s. It is divisible by 6 precisely when r + s = 6. Thus 8 + 10 = 18 makes the claim false, while 9 + 11 = 20 makes it true.(iv) Always true. 6a + 9b = 3(2a + 3b), so the sum always has a factor 3.(v) Sometimes true. 6a + 3b = 3(2a + b), divisible by 9 exactly when 2a + b is divisible by 3. Example: 18 + 9 = 27. Non-example: 12 + 9 = 21.

4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

SOLUTION If a number x leaves remainder 2 on division by both 3 and 4, then x − 2 is divisible by both 3 and 4, hence by LCM(3, 4) = 12. So x − 2 = 12n, i.e. x = 12n + 2. A few such numbers: 2, 14, 26, 38, 50, … (n = 0, 1, 2, …) (each leaves remainder 2 on division by 3 and by 4). ✓

5. “I hold some pebbles, not too many,
When I group them in 3’s, one stays with me.
Try pairing them up — it simply won’t do,
A stubborn odd pebble remains in my view.
Group them by 5, yet one’s still around,
But grouping by seven, perfection is found.
More than one hundred would be far too bold,
Can you tell me the number of pebbles I hold?”

SOLUTION The number leaves remainder 1 when grouped by 3, by 2 and by 5, so it is 1 more than a multiple of LCM(3, 2, 5) = 30. The positive candidates below 100 are 1, 31, 61 and 91; 1 is not divisible by 7. It must also be a multiple of 7 (“grouping by seven, perfection is found”). Of 31, 61, 91, only 91 = 7 × 13 is a multiple of 7. So the number of pebbles is 91. ✓
Ninety-one pebbles are arranged in thirteen rows of seven. Dividing by two, three or five leaves one pebble; dividing by seven leaves none.
Page 122, Question 5: candidates are 1, 31, 61 and 91; only 91 is divisible by 7.

6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

SOLUTION Let the three numbers be 6a + 2, 6b + 2 and 6c + 2. Sum = 6a + 6b + 6c + 6 = 6(a + b + c + 1). This is always a multiple of 6, so yes, Tathagat’s claim is true. (The three remainders 2 + 2 + 2 = 6 themselves form a multiple of 6.) ✓

7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. (i) 4779 + 661 (ii) 4779 − 661

SOLUTION Write 4779 = 7p + 5 and 661 = 7q + 3. (i) 4779 + 661 = 7p + 7q + 8 = 7(p + q + 1) + 1, so the remainder is 1 (the remainders 5 + 3 = 8 leave 1 on division by 7). (ii) 4779 − 661 = 7p + 5 − 7q − 3 = 7(p − q) + 2, so the remainder is 2 (5 − 3 = 2). Visually: draw each number as full rows of 7 with leftover tokens (3 and 5). Combining the leftovers (8) makes one more group of 7 with 1 left over for the sum; for the difference, 5 leftover minus 3 leftover gives 2.
Eight leftover tokens, five blue and three orange, regroup into seven plus one. Removing three from five leaves two tokens.
Page 123, Question 7: complete groups of 7 do not change the remainder.

8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

SOLUTION In each case the remainder is exactly 1 less than the divisor (3−1, 4−1, 5−1). So the number is 1 less than a common multiple of 3, 4 and 5. The smallest common multiple is LCM(3, 4, 5) = 60, so the smallest such number is 60 − 1 = 59. Why smallest: a number that is “1 less” than a multiple of all three divisors must be 1 less than a common multiple; 60 is the least common multiple, so 59 is the least such number. (Check: 59 = 3×19 + 2 = 4×14 + 3 = 5×11 + 4.) ✓
Remainders 2, 3 and 4 are each one short of divisors 3, 4 and 5. Adding one to 59 completes each group.
Page 123, Question 8: the blue tokens are the remainder; the orange token is the one added to reach 60.

Figure it Out (Page 126) — Divisibility by 9

1. Find, without dividing, whether the following numbers are divisible by 9. (i) 123   (ii) 405   (iii) 8888   (iv) 93547   (v) 358095

SOLUTION Add the digits and test divisibility by 9: (i) 1+2+3 = 6 → not divisible. (ii) 4+0+5 = 9 → divisible. (iii) 8+8+8+8 = 32 → not divisible. (iv) 9+3+5+4+7 = 28 → not divisible. (v) 3+5+8+0+9+5 = 30 → not divisible. So only 405 (ii) is divisible by 9. ✓

2. Find the smallest multiple of 9 with no odd digits.

SOLUTIONThe intended answer is the smallest positive multiple: 288. Its digits are all even and 2 + 8 + 8 = 18, so it is divisible by 9.Why no smaller positive answer? A sum of even digits is even. A positive digit-sum divisible by 9 must therefore be at least 18. One or two even digits have sum at most 8 or 16, so at least three digits are needed.For the smallest three-digit number, choose the smallest nonzero even hundreds digit, 2. The remaining sum must be 16, forcing the tens and ones to be 8 and 8. Thus 288 = 9 × 32.If zero is allowed, 0 itself is a multiple of 9 with no odd digit. The textbook’s answer uses the positive-number interpretation.

3. Find the multiple of 9 that is closest to the number 6000.

SOLUTION 6000 ÷ 9 = 666.67, so 9 × 666 = 5994 and 9 × 667 = 6003. Distances: 6000 − 5994 = 6 and 6003 − 6000 = 3. The closer one is 6003.

4. How many multiples of 9 are there between the numbers 4300 and 4400?

SOLUTION First multiple > 4300: 9 × 478 = 4302. Last multiple < 4400: 9 × 488 = 4392. Count = 488 − 478 + 1 = 11 multiples (4302, 4311, …, 4392). ✓

Figure it Out (Page 131, Digital Roots) — Solutions

1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

SOLUTION Digital root behaves like “add modulo 9.” Adding 10 adds 1 + 0 = 1 to the digital root. New digital root = 5 + 1 = 6. (For example, the 8-digit number 40000001 has digital root 5, and 40000011 has digital root 6.) ✓

2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

SOLUTION Take 10 and keep adding 11: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, … Their digital roots are 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, … Observation: since 11 adds 1 + 1 = 2 to the digital root each time, the digital roots go up by 2 (mod 9). They cycle through every value 1–9 with period 9, in the order 1, 3, 5, 7, 9, 2, 4, 6, 8 — then repeat. ✓
A circular cycle of digital roots runs 1, 3, 5, 7, 9, 2, 4, 6, 8 and returns to 1 when eleven is repeatedly added.
Page 131, Question 2: the cycle repeats after nine steps.

3. What will be the digital root of the number 9a + 36b + 13?

SOLUTION 9a + 36b + 13 = 9a + 36b + 9 + 4 = 9(a + 4b + 1) + 4. 9(a + 4b + 1) is a multiple of 9 (digital root 9), so the whole number is 4 more than a multiple of 9. Hence the remainder on division by 9 is 4, so the digital root is 4. ✓

4. Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root. (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

SOLUTION (i) No consistent pattern. An even number can have an odd or even digital root (12 → 3 odd, 24 → 6 even), and likewise for odd numbers. Parity and digital root are independent. (ii) Division by 3: digital roots 3, 6, 9 → remainder 0; roots 1, 4, 7 → remainder 1; roots 2, 5, 8 → remainder 2. Division by 9: the remainder equals the digital root, except a digital root of 9 means remainder 0.

Figure it Out (Page 132, Digits in Disguise) — Solutions

1. 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.

SOLUTION Digit-sum = 3 + 1 + z + 5 = 9 + z, which must be a multiple of 9. 9 + z = 9 gives z = 0; 9 + z = 18 gives z = 9. Both are single digits. So z = 0 or z = 9. There are two answers because 9 + z can be 9 or 18 while z stays a valid digit. ✓

2. “I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.

SOLUTION Let the numbers be a = 12n + 8 and b = 12m − 4. Sum = 12(n + m) + 4 = 12k + 4. This is always a multiple of 4 but not always a multiple of 8: e.g. k = 2 gives 28, which is not divisible by 8. So Snehal’s claim is false. ✓

3. When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

SOLUTION Let the two multiples of 3 be 3m and 3n. Their sum is 3m + 3n = 3(m + n). This is a multiple of 6 exactly when (m + n) is even (so that 3(m+n) has a factor of 2 as well). Cases: both m, n even or both odd → m+n even → sum is a multiple of 6 (e.g. 6 + 12 = 18). One even and one odd → m+n odd → sum is a multiple of 3 but not 6 (e.g. 3 + 6 = 9). ✓

4. 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) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

SOLUTION (i) Yes, always true. Reversing the digits does not change the digit-sum, and divisibility by 9 depends only on the digit-sum. (ii) Yes — any rearrangement (shuffle) of the digits keeps the same digit-sum, so the new number is still a multiple of 9. ✓

5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

SOLUTION A multiple of 18 must be divisible by 2 and by 9. For divisibility by 2, the units digit b is even: b ∈ {0, 2, 4, 6, 8}. For divisibility by 9, the digit-sum 4 + 8 + a + 2 + 3 + b = 17 + a + b must be a multiple of 9, i.e. a + b = 1 or 10 (since a + b ranges 0–17). Pairing each even b with the a that makes a + b = 1 or 10 (with a a digit): (a, b) = (1, 0), (8, 2), (6, 4), (4, 6), (2, 8).

6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

SOLUTION 44 = 4 × 11, so the number must be divisible by 4 and by 11. By 4: the last two digits “q8” must be a multiple of 4 → q8 ∈ {08, 28, 48, 68, 88} → q ∈ {0, 2, 4, 6, 8}. By 11: (sum of digits at odd places) − (sum at even places) = (8 + 7 + 3) − (q + p) = 18 − (p + q) must be 0 or a multiple of 11. Because p ≤ 9 and even q ≤ 8, p + q ≤ 17. The difference 18 − (p + q) lies from 1 to 18, so its only possible multiple of 11 is 11. Hence p + q = 7. Combining q even with p + q = 7: (p, q) = (7, 0), (5, 2), (3, 4), (1, 6).

7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

SOLUTION One set is 2, 3, 4 (2 is a multiple of 2, 3 of 3, 4 of 4). Yes, there are more. The pattern repeats every LCM(2, 3, 4) = 12 starting positions, so the next set is 14, 15, 16, then 26, 27, 28, and so on. They occur once in every block of 12 consecutive numbers. ✓

8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

SOLUTION45000 = 36 × 1250. To stay strictly between the two endpoints, start with the next multiple and add 36 each time.45036, 45072, 45108, 45144, 45180.These equal 36 × 1251 through 36 × 1255, and all lie strictly between 45000 and 47000.
A number line lists 45036, 45072, 45108, 45144 and 45180 at increments of 36.
Question 8: 45000 is an endpoint, so choose five numbers strictly inside the stated interval.

9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

SOLUTION Consecutive even numbers differ by 2. Since 5p is even and p is an integer, p must be even. With middle term 5p, the two below are 5p − 2 and 5p − 4, and the two above are 5p + 2 and 5p + 4. In order: 5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4. So the other four are 5p − 4, 5p − 2, 5p + 2 and 5p + 4.

10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.

SOLUTION Divisible by 15 means divisible by 3 and 5, so the units digit is 0 or 5. To keep both the original and its reversal six-digit, choose a number ending in 5. If a shorter reversal is allowed, numbers ending in 0 can also qualify. The reversed number must be divisible by 6 (by 2 and 3), so it must be even — hence the original number’s first digit must be even (2, 4, 6 or 8). One valid choice: 200055. Check: 200055 = 15 × 13337 (digit-sum 12 is a multiple of 3, ends in 5); reversed it is 550002, which is even with digit-sum 12, so divisible by 6. ✓

11. 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; explain your conclusion.

SOLUTION Let n be any multiple of 11, so n = 11k. Doubling: 2n = 22k = 11 × (2k). This is always a multiple of 11. So Deepak’s conjecture is false — doubling any multiple of 11 gives a multiple of 11. ✓

12. Determine whether the statements below 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 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.

SOLUTION (i) Always true. 6x × 3y = 18xy = 9(2xy), a multiple of 9. (ii) Always true. Three consecutive even numbers are 2k, 2k+2, 2k+4; their sum is 6k + 6 = 6(k + 1), a multiple of 6. (iii) Always true. Swapping a↔b and c↔d keeps the same units digit (still even) and the same digit-sum (so still a multiple of 3); hence still a multiple of 6. (iv) Never true. 8(7b − 3) − 4(11b + 1) = 56b − 24 − 44b − 4 = 12b − 28 = 12(b − 2) − 4, which is 4 short of a multiple of 12 — never a multiple of 12.

13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

SOLUTION On division by 3, every number leaves remainder 0, 1 or 2. The sum is divisible by 3 exactly when the sum of the three remainders is a multiple of 3. Case 1 — all three remainders equal (0,0,0 / 1,1,1 / 2,2,2): the remainder sum is 0, 3 or 6 → sum divisible by 3. Case 2 — all three remainders different (0,1,2 in any order): sum of remainders = 3 → sum divisible by 3. Generalisation: the sum of three numbers is divisible by 3 iff their remainders mod 3 are either all the same or all different. ✓

14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of three consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

SOLUTIONTwo: one is even, so their product is divisible by 2.Three: at least one is even and at least one is divisible by 3. Since 2 and 3 are coprime, the product is divisible by 6.Four: there are two even integers. One is divisible by 4, and the other supplies another factor of 2, so the product contains a factor 8. There is also a factor 3, giving divisibility by 8 × 3 = 24. This avoids counting the same factor of 2 twice.Five: a block of four supplies a factor 24, and the five-number block contains a multiple of 5. Since 24 and 5 are coprime, the product is divisible by 120. If the block includes 0, the product is 0 and the divisibility statements still hold.

15. Solve the cryptarithms — (i) EF × E = GGG (ii) WOW × 5 = MEOW

SOLUTION(i) EF × E = GGG. GGG = 111G = 3 × 37 × G. Since the one-digit E is not divisible by 37, the two-digit EF must be divisible by 37. The only candidates are 37 and 74. The first gives 37 × 3 = 111; the second gives 74 × 7 = 518, which is not a repeated-digit number. Thus E = 3, F = 7, G = 1.(ii) WOW × 5 = MEOW. The units column requires 5W to end in W. Since W is a nonzero leading digit, W = 5 and the carry is 2. The tens column requires 5O + 2 to end in O, so 4O + 2 is divisible by 10: O = 2 or 7.If O = 2, then 525 × 5 = 2625 would make M = O = 2, forbidden for different letters. If O = 7, 575 × 5 = 2875, which works. Therefore W = 5, O = 7, M = 2, E = 8.
EF times E equals GGG becomes 37 times 3 equals 111. WOW times 5 equals MEOW becomes 575 times 5 equals 2875.
Question 15: E = 3, F = 7, G = 1 in (i); W = 5, O = 7, M = 2, E = 8 in (ii).

16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

SOLUTION Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4 (since 32 = 8 × 4 and 8 = 4 × 2). So the sets are nested: multiples of 32 inside multiples of 8 inside multiples of 4. This is option (iv). ✓
Nested regions place multiples of 32 inside multiples of 8 inside multiples of 4. Example numbers distinguish the outer regions.
Question 16: choose option (iv), the diagram with these nested sets.

Math Talk, divisibility tables and extra puzzles

Consecutive sums and parity (Pages 112–115)

SOLUTIONUsing at least two positive consecutive integers, not every natural number can be written this way: 1, 2 and 4 are small counterexamples. An odd number 2k + 1 ≥ 3 equals k + (k + 1). Some even numbers work, such as 6 = 1 + 2 + 3 and 12 = 3 + 4 + 5. The number 15 has several representations: 7 + 8, 4 + 5 + 6, and 1 + 2 + 3 + 4 + 5.Allowing negative integers changes the problem: 0 = −1 + 0 + 1. For any positive integer N, the consecutive integers from −N + 1 to N sum to N, because all opposite pairs cancel except N.For n, n + 1, n + 2, n + 3, the eight sign patterns +++, ++−, +−+, +−−, −++, −+−, −−+, −−− give 4n + 6, 2n, 2n + 2, −4, 2n + 4, −2, 0 and −2n − 6. They are all even. The results −4, −2 and 0 occur for every starting n.For any fixed list of integers, changing +b to −b changes the total by 2b. Thus all sign choices have the same parity. This works for any number of terms, not just four.
The eight expressions made from 3, 4, 5 and 6 have values 18, 6, 8, minus 4, 10, minus 2, zero and minus 12; all are even.
Pages 112–115: each of the three gaps has two choices, so there are 2³ = 8 expressions.

Breaking Even and pairs that make fours (Pages 115–118)

SOLUTIONEven arithmetic expressions: 43 + 37, 672 − 348, 4 × 347 × 3 and 543 − 479. The others shown—708 − 477, 809 + 214, 119 × 303 and 513³—are odd.Always even algebraic expressions: 2a + 2b = 2(a + b), 4m + 2n = 2(2m + n), 2u − 4v = 2(u − 2v), 13k − 5k = 8k, and 4k × 3j = 12kj.Not always even: 3g + 5h is even exactly when g and h have the same parity; 6m − 3n is even exactly when n is even; b² + 1 is even exactly when b is odd; x² + 2 is even exactly when x is even.Two even numbers are either 4p or 4p + 2. Their sum is divisible by 4 when both have remainder 0, or both have remainder 2. Mixed remainders 0 and 2 give remainder 2: for example, 8 + 6 = 14.

Remainder forms and divisibility proofs (Pages 121–127)

SOLUTIONRemainder 3 on division by 5 can be written 5k + 3 for k ≥ 0 or 5k − 2 for k ≥ 1. Both generate 3, 8, 13, 18, … . To represent an arbitrary even and an arbitrary odd number independently, use 2n and 2m + 1; using the same n would restrict them to consecutive numbers.For any whole number, write N = 10q + a. Since 10q is divisible by 2 and 5, these tests depend only on the last digit a. Likewise, N = 100q + r gives the last-two-digit test for 4, and N = 1000q + r gives the last-three-digit test for 8.Every decimal place value is 1 more than a multiple of 9, and also 1 more than a multiple of 3. Thus a number and its digit sum have the same remainder for either divisor. All four “if” statements on pages 125–126 are true: divisibility of the number and its digit sum are equivalent.
427 is split into four hundreds, two tens and seven ones. Removing multiples of nine leaves 4 plus 2 plus 7 equals 13, which leaves remainder four.
Each decimal place value is one more than a multiple of 9. The same argument works for 3.

Division by 11 and the completed table (Pages 128–129)

SOLUTIONStart with + at the units digit and alternate signs moving left. Reduce the resulting signed sum to a remainder from 0 to 10. For example, 462 gives 2 − 6 + 4 = 0, so it is divisible by 11. The two printed examples 320185 and 328105 both give remainder 8, although their signed sums differ by 11.
NumberAlternating sumRemainder ÷11
1588 − 5 + 1 = 44
8411 − 4 + 8 = 55
4811 − 8 + 4 = −38
55299 − 2 + 5 − 5 = 77
909044 − 0 + 9 − 0 + 9 = 220
8570766 − 7 + 0 − 7 + 5 − 8 = −110
The alternating sum of 328105 from the units digit is negative three. Adding eleven gives the actual remainder eight.
If the alternating sum is a multiple of 11, the remainder is 0. For other values, reduce it to a number from 0 to 10.
Number÷2÷3÷4÷5÷6÷8÷9÷10÷11
128YesNoYesNoNoYesNoNoNo
990YesYesNoYesYesNoYesYesYes
1586YesNoNoNoNoNoNoNoNo
275NoNoNoYesNoNoNoNoYes
6686YesNoNoNoNoNoNoNoNo
639210YesYesNoYesYesNoNoYesYes
429714YesYesNoNoYesNoYesNoNo
2856YesYesYesNoYesYesNoNoNo
3060YesYesYesYesYesNoYesYesNo
406839NoYesNoNoNoNoNoNoNo

In particular, 128 is divisible by 4: 128 = 4 × 32. Use the arithmetic when a prefilled textbook cell is inconsistent.

Combining divisibility tests (Pages 129–130)

SOLUTIONFor 6, check both 2 and 3. Among 38, 225, 186 and 64, only 186 passes both. It equals 6 × 31.For 24 = 2³ × 3, checking 8 and 3 supplies all required prime factors. Checking 4 and 6 only guarantees 2² × 3 = 12; it does not guarantee the extra factor of 2. For example, 12 passes 4 and 6 but fails 24.

Digital-root explorations (Page 130)

489710 reduces to 29, then 11, then 2, with remainder two on division by nine. 405 reduces to nine, but its remainder is zero.
For a positive number, digital roots 1–8 equal the remainder; digital root 9 means remainder 0.
SOLUTIONStrictly between 600 and 700: root 5: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698. Root 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691. Root 3: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696. Each list increases by 9.For twelve consecutive positive numbers 1–12, roots are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3. Positive multiples of 3 cycle 3, 6, 9; of 4 cycle 4, 8, 3, 7, 2, 6, 1, 5, 9; of 6 cycle 6, 3, 9. Numbers 1 more than nonnegative multiples of 6 cycle 1, 7, 4.Riddle: nine copies of the smallest odd digit, 1, give 111111111 = 11,11,11,111. Its number of digits, digit sum and digital root are all 9. Its Indian number name is eleven crore eleven lakh eleven thousand one hundred eleven.

Additional cryptarithms (Pages 131–132)

SOLUTIONWork column by column, keeping equal letters equal and different letters distinct. A printed digit is a fixed value; the distinct-letter rule applies between letters.
PuzzleValid numerical solution(s)
A1 + 1B = B071 + 19 = 90
AB + 37 = 6A25 + 37 = 62
ON + ON + ON = PO17 + 17 + 17 = 51; 24 + 24 + 24 = 72; 31 + 31 + 31 = 93
QR + QR + QR = PRR85 + 85 + 85 = 255
PQ × 8 = RS12 × 8 = 96
GH × H = 9K24 × 4 = 96 (from the listed options)
BYE × 6 = RAY105 × 6 = 630
UT × 3 = PUT50 × 3 = 150
AB × 5 = BC19 × 5 = 95
L2N × 2 = 2NP124 × 2 = 248; 125 × 2 = 250
XY × 4 = ZX23 × 4 = 92
PP × QQ = PRP22 × 11 = 242; 33 × 11 = 363; 44 × 11 = 484
JK × 6 = KKK74 × 6 = 444
For BYE × 6, a three-digit product forces B = 1 and Y ≤ 6. The last digit Y must be even. Column checks with distinct letters give 105 × 6 = 630. For PQ × 8, values 13 and above give at least three digits, while 10 and 11 violate the distinct-letter requirements.

Navakankari (Page 135)

Use the board below for the textbook’s two-player game. Each player starts with nine pawns and takes turns placing them on empty marked points. After placement, move a pawn to an adjacent empty point along a line. Forming a horizontal or vertical line of three allows a capture under the textbook’s rule: remove an opposing pawn that is outside an opposing line of three. A player wins when the opponent has fewer than three pawns or cannot move. Discuss the counting and planning used in choosing moves.

Three nested squares are connected at their side midpoints, making 24 marked playing positions.
Page 135: a line of three of your pawns forms a mill. Follow the textbook’s movement and capture rules.

Common Mistakes to Avoid

Watch out for these

  • Confusing “always / sometimes / never true” — one counterexample is enough to rule out “always,” but you need a general (algebraic) argument to prove “always.”
  • For divisibility by 9 or 3, testing the units digit instead of the digit-sum.
  • In the rule for 11, forgetting that the alternating sum can be 0 or any multiple of 11 (not just 0).
  • Thinking divisibility by a × b follows from divisibility by a and b when they share factors (e.g. 4 and 6 do not test 24; use 3 and 8 instead).
  • Writing the digital root as the divisor when the number is a multiple of 9 — the digital root is 9 but the remainder is 0.
  • Allowing a leading 0 or letting two different letters equal the same digit in a cryptarithm.

Practice MCQs & Assertion–Reason

1. The sum of two odd numbers is always:

(a) odd    (b) even    (c) a multiple of 3    (d) prime

2. A number is divisible by 9 if and only if:

(a) its units digit is 9    (b) its last two digits are divisible by 9    (c) the sum of its digits is divisible by 9    (d) it is even

3. The digital root of 489710 is:

(a) 2    (b) 5    (c) 9    (d) 11

4. Which number is divisible by 11?

(a) 158    (b) 481    (c) 90904    (d) 841

5. The remainder when 7309 is divided by 9 is:

(a) 0    (b) 1    (c) 7    (d) 9

6. Numbers that leave remainder 3 when divided by 5 are of the form:

(a) 3k + 5    (b) 5k + 3    (c) 5k − 3    (d) 3k

7. The product of any three consecutive integers is always a multiple of:

(a) 5    (b) 4    (c) 6    (d) 9

8. If 31z5 is a multiple of 9, the possible values of z are:

(a) 0 only    (b) 9 only    (c) 0 or 9    (d) 3 or 6

9. To check divisibility by 24, it is enough to check divisibility by:

(a) 4 and 6    (b) 3 and 8    (c) 2 and 12    (d) 6 and 12

10. The smallest number leaving remainders 2, 3, 4 when divided by 3, 4, 5 respectively is:

(a) 58    (b) 59    (c) 60    (d) 61

Answer key: 1-(b), 2-(c), 3-(a), 4-(c), 5-(b), 6-(b), 7-(c), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The number 405 is divisible by 9.

Reason: A number is divisible by 9 if the sum of its digits is divisible by 9.

A-R 2. Assertion: Doubling any multiple of 11 gives a multiple of 11.

Reason: If n = 11k, then 2n = 11(2k).

A-R 3. Assertion: A number divisible by both 4 and 6 must be divisible by 24.

Reason: A number divisible by both 4 and 6 must be divisible by LCM(4, 6) = 12.

A-R 4. Assertion: The digital root of every positive multiple of 9 is 9.

Reason: The remainder of any multiple of 9 on division by 9 is 0.

A-R 5. Assertion: The sum of three consecutive even numbers is always divisible by 6.

Reason: Three consecutive even numbers can be written as 2k, 2k+2, 2k+4, whose sum is 6(k+1).

Answer key: 1-(A), 2-(A), 3-(D), 4-(B), 5-(A).

Quick Revision Summary

  • Any combination a ± b ± c ± d of four numbers has a fixed parity, because a ± b always have the same parity.
  • If a divides M and N, it divides M + N and M − N; multiples of A stay divisible by A’s factors.
  • If A is divisible by k and m, then A is divisible by LCM(k, m).
  • Divisibility by 9 / 3 → test the digit-sum; digital root 9 means remainder 0; other positive digital roots equal the remainder.
  • Divisibility by 11 → alternating sum of digits is 0 or a multiple of 11.
  • Products of consecutive integers: 2 of them → multiple of 2; 3 → of 6; 4 → of 24; 5 → of 120.
  • Cryptarithms are solved by reasoning about leading digits, units digits and carrying.

How to score full marks in this chapter

For “always / sometimes / never” questions, always back “always true” with an algebraic proof and disprove with a clear counterexample. Show the digit-sum (for 3 and 9) or the alternating sum (for 11) explicitly. In word/remainder problems, convert each condition into mk + r form and use the LCM. For cryptarithms, state the constraint you used (range of the leading digit, units-digit match) before guessing.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 5 about?

Chapter 5, Number Play, explores parity (odd/even), sums of consecutive numbers, divisibility properties, the divisibility shortcuts for 3, 9 and 11 (with reasons), digital roots, and cryptarithms — with a strong focus on understanding why each rule works.

How many Figure it Out exercises are in Ganita Prakash Chapter 5?

There are four “Figure it Out” sets: on page 122 (8 questions), page 126 (4 questions), page 131 in Digital Roots (4 questions) and page 132 in Digits in Disguise (16 questions) — 32 questions in all, every one solved on this page.

What is the divisibility rule for 11 used in this chapter?

Place alternating + and − signs before the digits starting from the units place and add them up. If the result is 0 or a multiple of 11, the number is divisible by 11; otherwise reduce the signed sum modulo 11 to obtain a remainder from 0 to 10.

Are these Class 8 Maths Ganita Prakash Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for 2026–27.

Textbook reference: NCERT Ganita Prakash, Chapter 5 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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