NCERT Solutions for Class 8 Maths Chapter 5: Number Play – Free PDF Download

Chapter 5 — Number Play — explores number theory through puzzles: parity of algebraic expressions, sums of consecutive numbers, divisibility rules (9, 11, 18, 44), digital roots, remainder/modular reasoning, and cryptarithms (letter-for-digit puzzles). The chapter has four Figure It Out blocks totalling 32 questions. Below are complete, verified answers, sourced directly from the official NCERT PDF (including its own answer-key appendix) and independently recalculated. These Class 8 Mathematics Chapter 5 solutions are also useful as quick revision notes before exams.

NCERT Solutions for Class 8 Maths Chapter 5: Number Play

Figure It Out — Section 5.1 (8 Questions)

1. The sum of four consecutive numbers is 34. What are they?
7, 8, 9, 10.

2. If p is the greatest of five consecutive numbers, describe the other four in terms of p.
p−1, p−2, p−3, p−4.

3. Always/Sometimes/Never true, with algebraic justification: (i) sum of two even numbers is a multiple of 3 (ii) not divisible by 18 ⇒ not divisible by 9 (iii) two numbers not divisible by 6 ⇒ their sum not divisible by 6 (iv) sum of a multiple of 6 and a multiple of 9 is a multiple of 3 (v) sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
(i) Sometimes (e.g. 2+4=6 yes, 2+6=8 no). (ii) Sometimes (30 confirms, 27 refutes). (iii) Sometimes (9+11=20 no, 8+10=18 yes). (iv) Always true (6x+9y=3(2x+3y)). (v) Sometimes (18+9=27 yes, 12+9=21 no).

4. Find numbers leaving remainder 2 when divided by both 3 and 4. Give a general expression.
12n+2 (e.g. 14, 26, 38…).

5. Pebble riddle: remainder 1 mod 3, remainder 1 mod 5, remainder 0 mod 7, fewer than 100.
91.

6. Tathagat claims: any three numbers leaving remainder 2 mod 6, summed, always give a multiple of 6. True?
Yes, always true — (6a+2)+(6b+2)+(6c+2)=6(a+b+c+1).

7. 661÷7 leaves remainder 3, 4779÷7 leaves remainder 5. Without dividing, find the remainder of (i) their sum (ii) their difference.
(i) remainder 1 (3+5=8=7+1). (ii) remainder 2 (5−3=2).

8. Smallest number leaving remainder 2 mod 3, 3 mod 4, 4 mod 5.
Each remainder is 1 less than its divisor, so the answer is LCM(3,4,5)−1 = 59.

Figure It Out — Section 5.2 (4 Questions)

1. Without dividing, check divisibility by 9: 123, 405, 8888, 93547, 358095.
Only 405 is divisible by 9 (digit sums 6, 9, 32, 28, 30 — only 9 works).

2. Smallest multiple of 9 with no odd digits.
288 (digits 2,8,8 sum to 18).

3. Multiple of 9 closest to 6000.
6003 (667×9).

4. How many multiples of 9 lie between 4300 and 4400?
11.

Figure It Out — Section 5.3, Digital Roots (4 Questions)

1. An 8-digit number has digital root 5. What is the digital root of that number + 10?
6.

2. Start at any number and repeatedly add 11 — what pattern do the digital roots follow?
The digital roots cycle through all values 1–9, advancing by 2 each time (mod 9) — e.g. 1,3,5,7,9,2,4,6,8,1,3…

3. Digital root of 9a+36b+13?
= 9(a+4b+1)+4, so the digital root is always 4.

4. Is there a pattern between (i) parity and digital root, (ii) digital root and remainder mod 3 or mod 9?
(i) No consistent pattern. (ii) Digital roots 1,4,7 → remainder 1 mod 3; 2,5,8 → remainder 2 mod 3; 3,6,9 → remainder 0 mod 3. For mod 9: digital root = remainder (except digital root 9 → remainder 0).

Figure It Out — Section 5.4, Cryptarithms & More (16 Questions)

1. If 31z5 is a multiple of 9, find z. Why two answers?
z=0 or z=9 — because the digit sum without z is already 9, so both 9 and the next multiple (18) work.

2. Snehal claims: (12n+8) + (12m−4) is always a multiple of 8. True?
False — the sum simplifies to 12(n+m)+4, which is not generally a multiple of 8.

3. When is the sum of two multiples of 3 also a multiple of 6?
3m+3n=3(m+n); it’s a multiple of 6 only when (m+n) is even.

4. Sreelatha claims a multiple of 9, reversed, is still a multiple of 9. True? Any other shuffles?
True — digit sum is unchanged by reversal. Yes, any digit rearrangement works, since digit sum is invariant.

5. If 48a23b is a multiple of 18, list all (a,b) pairs.
(1,0), (8,2), (6,4), (4,6), (2,8) — b even (div by 2) and 17+a+b a multiple of 9 (div by 9).

6. If 3p7q8 is a multiple of 44, list all (p,q) pairs.
(7,0), (5,2), (3,4), (1,6) — from divisibility by 4 (automatic here) and by 11 (p+q=7).

7. Find three consecutive numbers where the 1st is a multiple of 2, 2nd of 3, 3rd of 4. More such triples?
2, 3, 4; the pattern repeats every LCM(2,3,4)=12, so next is 14, 15, 16, then 26, 27, 28, and so on.

8. Write five multiples of 36 between 45,000 and 47,000.
45036, 45072, 45108, 45144, 45180 (each 36 apart).

9. The middle of 5 consecutive even numbers is 5p. Express the other four.
5p−4, 5p−2, 5p+2, 5p+4.

10. Write a 6-digit number divisible by 15 that becomes divisible by 6 when reversed.
Example family: 200025, 200055, 202005, etc. (last digit 5 for div-by-15, first digit even so the reversed number is even, digit sum a multiple of 3).

11. Deepak claims only some multiples of 11, when doubled, stay multiples of 11. True?
False — 2×(11k)=11(2k) is always a multiple of 11, for every multiple of 11, not just some.

12. Always/Sometimes/Never true: (i) product of a multiple of 6 and a multiple of 3 is a multiple of 9 (ii) sum of 3 consecutive even numbers is divisible by 6 (iii) if abcdef is a multiple of 6, so is badcef (iv) 8(7b−3)−4(11b+1) is a multiple of 12.
(i) Always true (6x·3y=18xy). (ii) Always true. (iii) Always true (digit sum and last digit unchanged by swapping non-last digits). (iv) Never true — simplifies to 12b−28 ≡ 8 (mod 12) always.

13. When is the sum of any 3 chosen numbers divisible by 3?
When all three leave the same remainder mod 3, or their three remainders are 0, 1, 2 in some combination summing to a multiple of 3.

14. Is the product of 2, 3, 4, 5 consecutive integers always a multiple of 2, 6, 24, 120 respectively?
Yes to all — 2 consecutive: always ÷2; 3 consecutive: always ÷6; 4 consecutive: always ÷24; 5 consecutive: always ÷120.

15. Solve the cryptarithms: (i) EF × E = GGG (ii) WOW × 5 = MEOW.
(i) E=3, F=7, G=1 (37×3=111). (ii) W=5, O=7, M=2, E=8 (575×5=2875).

16. Which Venn diagram shows the relationship between multiples of 4, 8 and 32?
The fully nested diagram: multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4.

Selected Math Talk & Try This Highlights

Shortcut for divisibility by 11 (alternating digit-sum difference).
A number is divisible by 11 if the difference between the sum of digits in odd positions and even positions is 0 or a multiple of 11.

Which of 2a+2b, 3g+5h, 4m+2n, 2u−4v, 13k−5k, 6m−3n, x²+2, b²+1, 4k×3j are always even?
Always even: 2a+2b, 4m+2n, 2u−4v, 13k−5k (=8k), 4k×3j (=12kj). Not always even: 3g+5h, 6m−3n, x²+2, b²+1.

Why This Chapter Matters

This chapter builds algebraic proof skills (proving ‘always/sometimes/never true’ claims rather than just checking examples) and divisibility reasoning that underlie later algebra, number theory, and competitive-exam number puzzles.

More on This Chapter

Extra Questions (HOTS) | Revision Notes | Formulas Handbook | Class 8 Maths Book

Frequently Asked Questions

Do I need to memorise all the divisibility rules separately?
Most composite-number rules (18, 36, 44) can be built by combining the rules for their coprime factors (e.g., 18=2×9, 44=4×11) — you only need the core rules for 2, 3, 4, 5, 8, 9, 10, 11.