📘 Chapter 5: Number Play – Complete Question Bank

Grade 8 Mathematics

NCERT Ganita Prakash

Comprehensive Worksheet

Logical Reasoning Problem-Solving Analytical Thinking Pattern Recognition Algebraic Generalization

📝 Study Notes & Key Properties

🔢 Insert NCERT image from p. 84 here – Consecutive number sums

• 5.1 Sums of Consecutive Numbers: Every odd number = sum of two consecutive numbers.
• 5.2 Parity of Expressions: For any 4 consecutive numbers, all +/– expressions yield even numbers.
• 5.3 Divisibility by 4: Even numbers → multiples of 4 (remainder 0) or remainder 2.
• 5.4 Divisibility Rules: If a divides M & N → divides M+N and M–N.
• 5.5 Remainders: Numbers leaving remainder r when divided by n: nk + r.
• 5.6 Digital Roots: Sum digits repeatedly until single digit. Multiples of 9 → digital root 9.

🔢 Insert NCERT image from p. 86 here – Divisibility shortcuts table

🧮 Multiple Choice Questions (20)

1. Which of the following is NOT a sum of consecutive natural numbers?

(a) 7

(b) 10

(c) 11

(d) 15

Explanation: 11 cannot be expressed as a sum of consecutive natural numbers. Powers of 2 generally cannot, except 2 itself. 7=3+4, 10=1+2+3+4, 15=7+8=4+5+6.

2. For any 4 consecutive numbers, all expressions with ‘+’ and ‘–’ yield:

(a) Odd numbers

(b) Even numbers

(c) Prime numbers

(d) Multiples of 3

Explanation: Changing a sign changes the value by an even number. Starting expression (e.g., a+b-c-d) gives even result, so all 8 possibilities yield even numbers.

3. The sum of two even numbers is divisible by 4 if:

(a) Both are multiples of 4

(b) Both leave remainder 2

(c) Either (a) or (b)

(d) None

Explanation: Even numbers are either multiples of 4 (4k) or 4k+2. Sum of two same type: 4k+4m=4(k+m) or (4k+2)+(4m+2)=4(k+m+1).

4. If a number is divisible by 8, it is also divisible by:

(a) 2

(b) 4

(c) Both (a) and (b)

(d) Only 2

Explanation: 8 = 2³, so any multiple of 8 contains factor 2²=4 and 2¹=2.

5. The remainder when 427 is divided by 9 is:

(a) 4

(b) 5

(c) 6

(d) 7

Explanation: Digit sum: 4+2+7=13 → 1+3=4. For divisibility by 9, remainder = digital root (unless digital root is 9, remainder 0).

🔍 Assertion & Reasoning (20)

1. Assertion: All odd numbers can be expressed as sums of two consecutive numbers.
Reason: Odd numbers are of the form 2n+1, which equals n + (n+1).

(a) Both correct, Reason explains Assertion

(b) Both correct, Reason does not explain

(c) Assertion correct, Reason wrong

(d) Both wrong

Explanation: Assertion is true (e.g., 7=3+4, 9=4+5). Reason correctly shows odd number 2n+1 = n + (n+1), which are consecutive integers.

2. Assertion: For 4 consecutive numbers, all ‘+’/‘–’ expressions give even results.
Reason: Changing a sign changes value by an even number.

(a) Both correct, Reason explains

(b) Both correct, no relation

(c) Assertion correct, Reason wrong

(d) Both wrong

Explanation: Assertion true (verified). Reason: Switching +b to -b changes total by 2b (even), so parity remains same.

✅ True/False (10)

1. Every even number can be expressed as a sum of consecutive numbers.

True

False

Explanation: False. Some even numbers like 2, 8, 32 (powers of 2) cannot be expressed as sums of consecutive natural numbers.

2. Digital root of a multiple of 9 is always 9.

True

False

Explanation: True. Multiples of 9 have digit sums divisible by 9, and repeated summing yields 9 (except 0 which gives 9 as digital root).

📝 Short Answer I (2 Marks × 15)

1. Express 15 as sums of consecutive numbers in two ways.

Answer: 15 = 7 + 8
15 = 4 + 5 + 6
Also 15 = 1 + 2 + 3 + 4 + 5

2. Show that for 4 consecutive numbers, a ± b ± c ± d is always even.

Answer: Let numbers be n, n+1, n+2, n+3. Changing a sign changes total by even number (e.g., +b to -b changes by 2b). Starting expression n+(n+1)-(n+2)-(n+3) = -4 (even). All other expressions have same parity.

📝 Short Answer II (3 Marks × 10)

1. Prove: If a number is divisible by both 9 and 4, it is divisible by 36.

Proof: Let N be divisible by 9 → contains factor 3². Divisible by 4 → contains factor 2². Thus N contains 2²×3²=36 as factor.
Alternatively: LCM(9,4)=36. If divisible by both, divisible by LCM.

2. Solve cryptarithm: UT × 3 = PUT

Solution: Try T=7 → 7×3=21, carry 2. U×3+2 ends with U. Try U=5 → 5×3+2=17, doesn't end with 5. Try U=1 → 1×3+2=5, doesn't end with 1. Try U=7 → 7×3+2=23, ends with 3? No.
Actually: 17×3=51 → U=1, T=7, P=5 ✓.

📚 Long Answer (5 Marks × 10)

1. Explore and write which numbers can be expressed as sums of consecutive numbers in more than one way. Provide reasoning.

Answer:
• Numbers with odd divisors >1 can be expressed in multiple ways.
• Example: 15 (divisors: 1,3,5,15) → 3 ways.
• General: If N has an odd divisor d>1, it can be written as sum of d consecutive numbers centered at N/d.
• Powers of 2 have only one representation (as single number).
• Formula: Number of ways = number of odd divisors of N minus 1.
• Example: 45 has odd divisors 1,3,5,9,15,45 → 6-1=5 ways.

2. Prove that the sum of three consecutive even numbers is divisible by 6.

Proof: Let three consecutive even numbers be 2n, 2n+2, 2n+4.
Sum = 2n + (2n+2) + (2n+4) = 6n + 6 = 6(n+1).
Clearly divisible by 6 for all integer n.

🧩 Case-Based Questions

Case 1: Sums of Consecutive Numbers

Anshu is exploring sums of consecutive numbers. He writes: 7=3+4, 10=1+2+3+4, 12=3+4+5, 15=7+8=4+5+6=1+2+3+4+5. He wonders: Can every natural number be written as a sum of consecutive numbers?

1. Which of the following numbers CANNOT be expressed as a sum of consecutive natural numbers?

(a) 9

(b) 11

(c) 16

(d) 25

Explanation: 11 cannot be expressed. Powers of 2 generally cannot (except 2 itself). 9=4+5, 16= cannot, 25=12+13.

2. Which number can be expressed as a sum of consecutive numbers in the most ways?

(a) 15

(b) 21

(c) 30

(d) 45

Explanation: 45 can be written in 5 ways: 22+23, 14+15+16, 7+8+9+10+11, 5+6+7+8+9+10, 1+2+3+4+5+6+7+8+9.

3. Which statement is TRUE?

(a) All odd numbers can be expressed as sum of 2 consecutive numbers

(b) All even numbers can be expressed as sum of consecutive numbers

(c) 0 can be expressed using only positive consecutive numbers

(d) Prime numbers cannot be expressed as sum of consecutive numbers

Explanation: (a) True: odd number n = (n/2 - 0.5) + (n/2 + 0.5). (b) False: e.g., 2,8. (c) False: needs negative numbers. (d) False: e.g., 5=2+3.

4. How many ways can 18 be expressed as sum of consecutive natural numbers?

(a) 1

(b) 2

(c) 3

(d) 4

Explanation: 18 = 5+6+7 = 3+4+5+6. Two ways only.

Case 2: Parity and 4 Consecutive Numbers

Take any 4 consecutive numbers, say 3,4,5,6. Place '+' and '–' signs between them in all 8 possible ways. All results are even numbers.

1. What is common about all results?

(a) All are positive

(b) All are negative

(c) All are even numbers

(d) All are multiples of 3

Explanation: Verified by evaluating all 8 expressions: 18,6,8,-4,10,-2,0,-12 → all even.

2. Why are all results even?

(a) Because 4 consecutive numbers contain 2 evens and 2 odds

(b) Because sum of 4 consecutive numbers is always even

(c) Because changing a sign changes value by an even number

(d) Both (a) and (c)

Explanation: Both reasons contribute. The parity remains same because sign changes alter by even amounts.

🔢 Insert NCERT image from p. 90 here – Digital root example

❓ "Figure It Out" Questions

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

Solution: Let numbers be n, n+1, n+2, n+3.
Sum = 4n + 6 = 34 → 4n = 28 → n=7.
Numbers: 7, 8, 9, 10.

2. "I hold some pebbles... When grouped by 3's, one remains... by 5's, one remains... by 7's, none remain. Less than 100. How many pebbles?"

Solution: N mod 3=1, mod 5=1, mod 7=0, N<100.
N=7k, try multiples of 7: 7,14,21,28,35,42,49,56,63,70,77,84,91,98.
Check odd (since mod 2=1): 7,21,35,49,63,77,91.
Check mod 5=1: 21,91.
Check mod 3=1: 91 ✓.
Answer: 91 pebbles.

📚 Chapter 5: Number Play – Complete Question Bank with Answers & Explanations

NCERT Ganita Prakash | Grade 8 Mathematics | © Educational Use Only