Number Play is the study of number patterns using parity, factors, multiples, divisibility, remainders, and digit logic. A number statement becomes stronger when students test it with algebra, examples, and counterexamples.
A good number pattern is never accepted only because it works once. Important Questions Class 8 Maths Chapter 5 help students test patterns using parity, consecutive numbers, plus-minus expressions, factors, multiples, divisibility shortcuts, digital roots, remainders, and cryptarithms. The 2026 NCERT chapter Number Play in Ganita Prakash Class 8 trains students to prove claims through algebra, visual models, examples, and counterexamples.
Key Takeaways
- Parity: Even and odd behaviour helps predict sums, differences, and signed expressions.
- Divisibility: Rules for 3, 9, and 11 come from place-value remainders.
- Remainders: Numbers with fixed remainders can be written as algebraic forms like (5k + 3).
- Cryptarithms: Letter-digit puzzles use place value, multiplication facts, and divisibility logic.
Important Questions Class 8 Maths Chapter 5 Structure 2026
| Concept |
Rule |
Key Use |
| Parity |
Even ± even = even, odd ± odd = even, odd ± even = odd |
Signed expressions and consecutive numbers |
| Divisibility |
Digit sum for 3 and 9, alternate digit sum for 11 |
Quick checks without long division |
| Remainders |
Numbers leaving remainder r on division by n are (nk + r) |
Algebraic forms and number puzzles |
Number Play Class 8 Chapter Overview
Number Play Class 8 is about reasoning with numbers, not memorising shortcuts. It begins with consecutive numbers and shows how plus and minus signs affect parity.
The chapter then moves into multiples of 4, always-sometimes-never claims, divisibility by 3, 9, and 11, digital roots, and cryptarithms. Each topic asks students to justify why a pattern works.
Class 8 Maths Chapter 5 Important Questions On Parity
Parity means whether a number is even or odd. It helps students predict the result of an expression without evaluating every case.
The chapter uses four numbers and changing signs to show that parity can remain unchanged.
Q1. What Is Parity In Class 8 Maths?
Parity tells whether a number is even or odd.
Even numbers are divisible by 2. Odd numbers leave remainder 1 when divided by 2.
Example: 18 has even parity, while 23 has odd parity.
Q2. Why Do All Expressions (a \pm b \pm c \pm d) Have The Same Parity?
All expressions (a \pm b \pm c \pm d) have the same parity because changing one sign changes the value by an even number.
If (+b) changes to (-b), the difference between the two expressions is (2b). Since (2b) is always even, parity does not change.
Example: (a + b - c - d) and (a - b - c - d) differ by (2b).
Q3. Evaluate (3 + 4 - 5 + 6) And (3 - 4 - 5 - 6). What Do You Notice?
Both expressions give even numbers.
- First expression: (3 + 4 - 5 + 6)
- Value = (8)
- Second expression: (3 - 4 - 5 - 6)
- Value = (-12)
- Both (8) and (-12) are divisible by 2
Final Answer: Both results are even.
Q4. Is A Negative Number Like (-12) Even?
Yes, (-12) is an even number.
A number is even if it has 2 as a factor. Negative numbers can also have 2 as a factor.
Example: (-12 = 2 \times -6).
Q5. What Happens When One Sign Is Switched In (a + b - c - d)?
The value changes by an even number.
- Original expression: (a + b - c - d)
- Changed expression: (a - b - c - d)
- Difference = ((a + b - c - d) - (a - b - c - d))
- Difference = (2b)
- (2b) is always even
Final Answer: The parity remains the same.
Consecutive Numbers Class 8 Questions
Consecutive numbers differ by 1. They help students convert number patterns into algebraic expressions.
The chapter starts with sums like (7 = 3 + 4), (10 = 1 + 2 + 3 + 4), and (15 = 7 + 8).
Q6. The Sum Of Four Consecutive Numbers Is 34. Find The Numbers.
The four consecutive numbers are 7, 8, 9, and 10.
- Let the numbers be (x, x + 1, x + 2, x + 3).
- Their sum is 34.
- (x + x + 1 + x + 2 + x + 3 = 34)
- (4x + 6 = 34)
- (4x = 28)
- (x = 7)
Final Answer: 7, 8, 9, 10
Q7. If (p) Is The Greatest Of Five Consecutive Numbers, Write The Other Four.
The other four numbers are (p - 4, p - 3, p - 2,) and (p - 1).
The greatest number is (p). Each previous number is one less than the next.
Final sequence: (p - 4, p - 3, p - 2, p - 1, p)
Q8. Can Every Odd Number Be Written As A Sum Of Two Consecutive Numbers?
Yes, every odd number can be written as a sum of two consecutive numbers.
Two consecutive numbers can be written as (n) and (n + 1). Their sum is (2n + 1), which is odd.
Example: (15 = 7 + 8).
Q9. Can 10 Be Written As A Sum Of Consecutive Natural Numbers?
Yes, 10 can be written as (1 + 2 + 3 + 4).
The numbers 1, 2, 3, and 4 are consecutive natural numbers. Their sum is 10.
Calculation: (1 + 2 + 3 + 4 = 10).
Q10. Can 0 Be Written As A Sum Of Consecutive Integers?
Yes, 0 can be written as a sum of consecutive integers if negative numbers are allowed.
A positive and a negative number can cancel each other. This does not work with natural numbers only.
Example: (-1 + 0 + 1 = 0).
Algebraic Expressions Class 8 Maths Questions On Even Numbers
Algebra gives a reason behind number patterns. A single example can support a claim, but it cannot prove an “always” statement.
To prove an expression is always even, students should show 2 as a factor.
Q11. Which Algebraic Expressions Always Give Even Numbers?
Algebraic expressions with 2 as a factor always give even numbers.
If an expression can be written as (2 \times) an integer expression, it is always even. This works for all integer values of the variables.
Example: (2a + 2b = 2(a + b)).
Q12. Is (4m + 2n) Always Even?
Yes, (4m + 2n) is always even.
- Given expression: (4m + 2n)
- Take 2 common: (2(2m + n))
- The expression has 2 as a factor.
- Any integer multiplied by 2 is even.
Final Answer: Always even
Q13. Is (13k - 5k) Always Even?
Yes, (13k - 5k) is always even.
- Given expression: (13k - 5k)
- Simplify: (8k)
- Write (8k = 2(4k))
- The expression has 2 as a factor.
Final Answer: Always even
Q14. Is (x^2 + 2) Always Even?
No, (x^2 + 2) is not always even.
If (x) is even, (x^2) is even. If (x) is odd, (x^2) is odd.
Example: For (x = 6), (x^2 + 2 = 38). For (x = 3), (x^2 + 2 = 11).
Q15. Is (3g + 5h) Always Even?
No, (3g + 5h) is sometimes even.
The parity depends on (g) and (h). It is even when (g) and (h) have the same parity.
Example: (g = 1, h = 1) gives 8. (g = 1, h = 2) gives 13.
Multiples And Factors Class 8 Questions
Multiples and factors help test whether a number divides another number exactly. The chapter uses them to judge claims as always true, sometimes true, or never true.
Students should write one example and one counterexample where needed.
Q16. When Is The Sum Of Two Even Numbers A Multiple Of 4?
The sum is a multiple of 4 when both even numbers leave the same remainder on division by 4.
Two multiples of 4 add to a multiple of 4. Two even non-multiples of 4 also add to a multiple of 4.
Examples: (12 + 16 = 28), and (6 + 10 = 16).
Q17. What Happens When A Multiple Of 4 Is Added To An Even Number That Is Not A Multiple Of 4?
The result is not a multiple of 4.
- Multiple of 4 = (4p)
- Even non-multiple of 4 = (4q + 2)
- Sum = (4p + 4q + 2)
- Sum = (4(p + q) + 2)
Final Answer: The remainder is 2 on division by 4.
Q18. If 8 Divides Two Numbers Separately, Does 8 Divide Their Sum?
Yes, 8 divides their sum.
If the numbers are (8a) and (8b), their sum is (8a + 8b). This equals (8(a + b)).
Example: (16 + 56 = 72), and 72 is divisible by 8.
Q19. If 8 Divides Two Numbers Separately, Does 8 Divide Their Difference?
Yes, 8 divides their difference.
If the numbers are (8a) and (8b), their difference is (8a - 8b). This equals (8(a - b)).
Example: (80 - 120 = -40), and (-40) is divisible by 8.
Q20. If A Number Is Divisible By 12, Is It Divisible By All Factors Of 12?
Yes, a number divisible by 12 is divisible by all factors of 12.
A multiple of 12 contains all factor relationships of 12. The factors of 12 are 1, 2, 3, 4, 6, and 12.
Example: 48 is divisible by 2, 3, 4, 6, and 12.
Q21. If A Number Is Divisible By 7, Is It Divisible By Every Multiple Of 7?
No, this is only sometimes true.
A number divisible by 7 may or may not be divisible by a larger multiple of 7. The larger multiple must also divide the number.
Example: 42 is divisible by 14, but not by 28.
Always Sometimes Never Maths Class 8 Important Questions
Always-sometimes-never questions need proof, examples, and counterexamples. This is one of the most important reasoning styles in Class 8 Maths Chapter 5 Number Play.
A single counterexample is enough to disprove an “always true” claim.
Q22. Is The Sum Of Two Even Numbers Always A Multiple Of 3?
No, this statement is sometimes true.
Example: (2 + 4 = 6), which is a multiple of 3. But (2 + 6 = 8), which is not.
Final Answer: Sometimes true
Q23. If A Number Is Not Divisible By 18, Is It Also Not Divisible By 9?
No, this statement is sometimes true.
A number may be divisible by 9 but not by 18. This happens when the number is an odd multiple of 9.
Example: 27 is divisible by 9, but not by 18.
Q24. If Two Numbers Are Not Divisible By 6, Is Their Sum Not Divisible By 6?
No, this statement is sometimes true.
Two non-multiples of 6 can add to a multiple of 6. A counterexample proves the statement is not always true.
Example: (2 + 4 = 6).
Q25. Is The Sum Of A Multiple Of 6 And A Multiple Of 9 A Multiple Of 3?
Yes, this statement is always true.
A multiple of 6 has 3 as a factor. A multiple of 9 also has 3 as a factor.
Example: (12 + 18 = 30), and 30 is divisible by 3.
Q26. Is The Sum Of A Multiple Of 6 And A Multiple Of 3 Always A Multiple Of 9?
No, this statement is sometimes true.
Example: (6 + 3 = 9), which is a multiple of 9. But (12 + 3 = 15), which is not.
Final Answer: Sometimes true
Q27. If A Number Is Divisible By Both 9 And 4, Must It Be Divisible By 36?
Yes, this statement is always true.
The LCM of 9 and 4 is 36. If a number is divisible by both, it must contain all prime factors of 36.
Example: 72 is divisible by 9, 4, and 36.
Q28. If A Number Is Divisible By Both 6 And 4, Must It Be Divisible By 24?
No, this statement is sometimes true.
The LCM of 6 and 4 is 12, not 24. A number divisible by both 6 and 4 need only be divisible by 12.
Example: 12 is divisible by 6 and 4, but not by 24.
Remainder Questions Class 8 With Algebraic Forms
Remainders can be described neatly using algebra. A number that leaves remainder (r) when divided by (n) can be written as (nk + r).
The chapter uses this idea for expressions like (5k + 3) and (5k - 2).
Q29. Which Expression Shows Numbers That Leave Remainder 3 When Divided By 5?
The expression is (5k + 3).
Numbers divisible by 5 have the form (5k). Numbers leaving remainder 3 are 3 more than multiples of 5.
Examples: 3, 8, 13, 18, and 23.
Q30. Can (5k - 2) Also Show Numbers Leaving Remainder 3 When Divided By 5?
Yes, (5k - 2) can show such numbers when (k \geq 1).
It means 2 less than a multiple of 5. That gives the same remainder as 3 more than the previous multiple of 5.
Example: (5 \times 2 - 2 = 8).
Q31. Find Numbers That Leave Remainder 2 When Divided By 3 And 4.
The numbers have the form (12k + 2).
- The number leaves remainder 2 on division by 3.
- It also leaves remainder 2 on division by 4.
- It is 2 more than a common multiple of 3 and 4.
- LCM of 3 and 4 is 12.
Final Answer: (12k + 2), such as 2, 14, 26, 38
Q32. Tathagat Adds Three Numbers That Each Leave Remainder 2 When Divided By 6. Is The Sum A Multiple Of 6?
Yes, the sum is always a multiple of 6.
- Each number has the form (6k + 2).
- Three such numbers are (6a + 2), (6b + 2), and (6c + 2).
- Sum = (6a + 6b + 6c + 6)
- Sum = (6(a + b + c + 1))
Final Answer: Always a multiple of 6
Q33. What Remainder Does (4779 + 661) Leave When Divided By 7?
The expression leaves remainder 1 when divided by 7.
- 4779 leaves remainder 5 on division by 7.
- 661 leaves remainder 3 on division by 7.
- Remainder of sum = (5 + 3 = 8)
- (8) leaves remainder 1 on division by 7.
Final Answer: Remainder = 1
Q34. What Remainder Does (4779 - 661) Leave When Divided By 7?
The expression leaves remainder 2 when divided by 7.
- 4779 leaves remainder 5 on division by 7.
- 661 leaves remainder 3 on division by 7.
- Remainder of difference = (5 - 3 = 2)
Final Answer: Remainder = 2
Divisibility Rules Class 8 Questions
Divisibility shortcuts help students test numbers quickly. The chapter does not only give rules, it explains why they work through place value.
The rules for 3, 9, and 11 are central to Number Play Class 8.
Q35. What Is The Divisibility Rule For 9?
A number is divisible by 9 if the sum of its digits is divisible by 9.
This works because 10, 100, 1000, and higher place values each leave remainder 1 when divided by 9.
Example: 7309 has digit sum 19. Then (1 + 9 = 10), and (1 + 0 = 1).
Q36. Check Whether 358095 Is Divisible By 9.
358095 is not divisible by 9.
- Add the digits: (3 + 5 + 8 + 0 + 9 + 5)
- Sum = 30
- 30 is not divisible by 9.
- So 358095 is not divisible by 9.
Final Answer: Not divisible by 9
Q37. What Is The Divisibility Rule For 3?
A number is divisible by 3 if the sum of its digits is divisible by 3.
This rule works because each place value leaves remainder 1 when divided by 3. So the digit sum decides divisibility.
Example: 123 has digit sum 6, so 123 is divisible by 3.
Q38. Check Whether 93547 Is Divisible By 9.
93547 is not divisible by 9.
- Digit sum = (9 + 3 + 5 + 4 + 7)
- Digit sum = 28
- 28 is not divisible by 9.
Final Answer: Not divisible by 9
Q39. What Is The Divisibility Rule For 11?
A number is divisible by 11 if the difference between alternate digit sums is 0 or a multiple of 11.
Place values alternately behave as 1 more and 1 less than a multiple of 11. This creates the alternate digit sum rule.
Example: For 841, ((8 + 1) - 4 = 5). So 841 is not divisible by 11.
Q40. Check Whether 5529 Is Divisible By 11.
5529 is not divisible by 11.
- From the right, take alternate groups.
- First group: (9 + 5 = 14)
- Second group: (2 + 5 = 7)
- Difference = (14 - 7 = 7)
- 7 is not a multiple of 11.
Final Answer: Not divisible by 11
Q41. Why Does Checking Divisibility By 4 And 6 Not Prove Divisibility By 24?
Checking divisibility by 4 and 6 does not prove divisibility by 24 because their LCM is 12.
The factors 4 and 6 share a factor 2. So divisibility by both only guarantees divisibility by 12.
Example: 12 is divisible by 4 and 6, but not by 24.
Digital Root Class 8 Questions
Digital root is found by repeatedly adding digits until one digit remains. It helps track remainders when dividing by 9.
The chapter connects digital roots with older Indian methods of checking arithmetic.
Q42. What Is Digital Root Class 8?
Digital root is the single digit obtained by adding the digits repeatedly.
It follows the same remainder pattern as division by 9. Multiples of 9 have digital root 9.
Example: Digital root of 489710 is 2.
Q43. Find The Digital Root Of 489710.
The digital root is 2.
- Add digits: (4 + 8 + 9 + 7 + 1 + 0 = 29)
- Add again: (2 + 9 = 11)
- Add again: (1 + 1 = 2)
Final Answer: Digital root = 2
Q44. The Digital Root Of An 8-Digit Number Is 5. What Is The Digital Root Of 10 More Than That Number?
The new digital root is 6.
Adding 10 adds 1 more than a multiple of 9. So the digital root increases by 1.
Example: A digital root of 5 becomes 6 after adding 10.
Q45. What Is The Digital Root Of (9a + 36b + 13)?
The digital root is 4.
- (9a) is a multiple of 9.
- (36b) is also a multiple of 9.
- The expression has the same digital root as 13.
- (1 + 3 = 4)
Final Answer: Digital root = 4
Cryptarithms Class 8 Important Questions
Cryptarithms are digit puzzles where letters stand for digits. Different letters represent different digits.
The chapter uses cryptarithms to connect place value, multiplication, divisibility, and logical elimination.
Q46. What Are Cryptarithms In Class 8 Maths?
Cryptarithms are number puzzles where letters represent digits.
Each letter represents one digit only. The first digit of a number cannot be 0.
Example: In (A1 + 1B = B0), A and B must be digits.
Q47. If (PQ \times 8 = RS), Why Can (PQ) Not Be Greater Than 12?
(PQ) cannot be greater than 12 because the product would become a 3-digit number.
The product must be a 2-digit number (RS). Since (13 \times 8 = 104), any larger two-digit number fails.
Example: (12 \times 8 = 96) works as a 2-digit product.
Q48. Solve (GH \times H = 9K) From The Given Options.
One valid solution is (46 \times 2 = 92).
- (GH) is a two-digit number.
- (H) is also the units digit and multiplier.
- The product must be in the 90s.
- (46 \times 2 = 92).
- (G = 4, H = 6, K = 2).
Final Answer: (46 \times 2 = 92)
Q49. If (31z5) Is A Multiple Of 9, Find (z).
The value of (z) can be 0 or 9.
- Digit sum = (3 + 1 + z + 5)
- Digit sum = (9 + z)
- For divisibility by 9, (9 + z) must be a multiple of 9.
- Since (z) is a digit, (z = 0) or (z = 9).
Final Answer: (z = 0) or (z = 9)
Q50. If (48a23b) Is A Multiple Of 18, What Conditions Must (a) And (b) Satisfy?
The number must be divisible by both 2 and 9.
- Divisibility by 2 means (b) must be even.
- Digit sum = (4 + 8 + a + 2 + 3 + b = 17 + a + b)
- Divisibility by 9 means (17 + a + b) is a multiple of 9.
- Possible digit sums are 18 or 27.
Final Answer: (b) is even and (17 + a + b) is divisible by 9.
Class 8 Maths Important Questions Chapter-Wise