Class 8 : Maths – Lesson 5. Number Play

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EXPLANATION AND ANALYSIS

🎲 INTRODUCTION — PLAYING WITH NUMBERS

🔢 Numbers are not just symbols used for counting.
🧠 They follow patterns, rules, and relationships that make mathematics logical and interesting.

📘 This lesson focuses on:

🔵 recognising number patterns
🟡 understanding properties of numbers
🔴 using logic instead of lengthy calculations
🟣 enjoying mathematics through exploration

🧠 Number Play helps develop number sense, which is the heart of mathematics.

🔍 OBSERVING NUMBER PATTERNS
✨ Many numbers follow hidden patterns.
🧠 When we look carefully, we can predict results without heavy calculation.
📌 Examples of patterns include:
🔵 repeating digits
🟡 increasing and decreasing sequences
🔴 alternating numbers
🟣 symmetry in digits
🎯 Recognising patterns helps us:
🟢 simplify problems
🟠 solve faster
🟣 avoid mistakes
🔄 PLAYING WITH DIGITS OF NUMBERS

🔢 A number is made up of digits, and their position matters.

📍 Changing the order of digits can:

🔵 change the value of the number
🟡 affect divisibility
🔴 create predictable differences

🧠 Observing how digits behave builds confidence in handling numbers.

➕ ➖ EFFECT OF OPERATIONS ON NUMBERS

🧮 Numbers behave differently under different operations.

📌 Key ideas:

🔵 adding zero keeps a number unchanged
🟡 multiplying by one keeps the number same
🔴 multiplying by zero makes the result zero
🟣 subtraction depends on order
🟠 division is not always possible

🎯 Understanding these effects avoids careless errors.

🔢 EVEN AND ODD NUMBER PLAY

⚖️ Numbers can be classified as even or odd.

📘 Rules:

🔵 even numbers end in 0, 2, 4, 6, or 8
🔴 odd numbers end in 1, 3, 5, 7, or 9

🧠 Interesting observations:

🟢 even + even = even
🟡 odd + odd = even
🟣 even + odd = odd

🎲 Such patterns help solve problems mentally.

✖️ MULTIPLICATION PATTERNS
📐 Multiplication follows special patterns that make calculations easy.
✨ Examples:
🔵 multiplying by 10 adds a zero
🟡 multiplying by 100 adds two zeros
🔴 multiplying by 9 shows digit patterns
🟣 multiplying by 11 creates repeating sums
📌 These patterns reduce calculation time and improve accuracy.

🔢 DIVISIBILITY IDEAS

🧠 Divisibility rules help us check whether a number is divisible without dividing.

📘 Common ideas:

🔵 divisibility by 2 depends on last digit
🟡 divisibility by 5 depends on last digit
🔴 divisibility by 10 requires last digit zero

🎯 Using these rules saves time and avoids long division.

🔁 NUMBER REVERSAL AND RESULTS

🔄 Reversing digits of a number often creates interesting results.

📌 Observations include:

🔵 difference between a number and its reverse
🟡 sum of a number and its reverse
🟣 repeating patterns after reversal

🧠 Such activities sharpen logical thinking.

🧩 NUMBER TRICKS AND LOGIC

🎲 Some number problems look difficult but become easy when logic is applied.

📘 These tricks rely on:

🔵 fixed patterns
🟡 predictable outcomes
🟣 step-by-step reasoning

📌 The goal is not magic, but mathematical reasoning.

🔢 PLACE VALUE AND NUMBER BEHAVIOUR

📍 The value of a digit depends on its position.

🧠 Changing place value changes the entire number.

📘 Understanding place value helps in:

🔵 comparing numbers
🟡 estimating results
🔴 avoiding calculation mistakes

🧠 THINKING BEFORE CALCULATING

🚦 One key idea of Number Play is thinking before calculating.

📌 Instead of directly calculating, we should:

🔵 observe the pattern
🟡 predict the result
🟣 then verify

🎯 This approach improves speed and confidence.

🎯 REAL-LIFE CONNECTIONS OF NUMBER PLAY

🌍 Number patterns appear in:

🔵 calendars
🟡 clocks
🔴 phone numbers
🟣 PIN codes
🟠 seating arrangements

📘 Mathematics is not separate from life; it reflects daily patterns.

⚠️ COMMON ERRORS TO AVOID

🔴 ignoring digit position
🟡 applying operations blindly
🟣 forgetting number properties
🟠 rushing without observing patterns

✔️ Always pause and analyse before solving.

🌟 IMPORTANCE OF THIS LESSON

🏆 develops number sense
🟡 improves logical reasoning
🟣 makes calculations faster
🟠 builds base for algebra
🟢 encourages thinking over memorisation

🧾 SUMMARY

🔵 numbers follow patterns
🟡 digits behave differently based on position
🔴 operations affect numbers predictably
🟣 even–odd rules simplify problems
🟠 logic reduces calculation effort
🟢 number sense is essential

🔁 QUICK RECAP

🔵 observe patterns first
🟡 think before calculating
🟣 use logic confidently
🟠 apply shortcuts wisely
🔴 accuracy matters more than speed

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TEXTBOOK QUESTIONS

🔒 ❓ Q1. 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.
📌 ✅ Answer:
⬥ A number is divisible by 9 if the sum of its digits is divisible by 9.
⬥ Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z.
⬥ For divisibility by 9, 9 + z must be a multiple of 9.

🟢 Step 1 ⬥ Possible values:
⬥ 9 + z = 9 ⇒ z = 0
⬥ 9 + z = 18 ⇒ z = 9

⬥ Therefore, z = 0 or z = 9.
⬥ There are two answers because both values make the digit sum a multiple of 9.

🔒 ❓ Q2. “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.
📌 ✅ Answer:
⬥ Let the first number be 12a + 8.
⬥ Let the second number be 12b − 4.

🟢 Step 1 ⬥ Add the two numbers:
⬥ (12a + 8) + (12b − 4) = 12(a + b) + 4

⬥ 12(a + b) is divisible by 4, so the sum becomes 4 (3(a + b) + 1).
⬥ This is divisible by 4, but not always by 8.

➡️ Conclusion: Snehal’s claim is false. The sum is not always a multiple of 8.

🔒 ❓ Q3. 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.
📌 ✅ Answer:
⬥ Multiples of 3 can be written as 3k.

🟢 Step 1 ⬥ Case analysis:
⬥ 3(odd) + 3(odd) = 6(even) ⇒ divisible by 6
⬥ 3(even) + 3(even) = 6(even) ⇒ divisible by 6
⬥ 3(odd) + 3(even) = 3(odd) ⇒ not divisible by 6

⬥ A number is divisible by 6 if it is divisible by 2 and 3.

➡️ Generalisation:
⬥ The sum of two multiples of 3 is a multiple of 6 only when both are even multiples or both are odd multiples of 3.

🔒 ❓ Q4. 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.
📌 ✅ Answer:
⬥ Divisibility by 9 depends on the sum of digits, not their order.
⬥ Reversing digits does not change the digit sum.

➡️ Conclusion: The conjecture is true for all multiples of 9.

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
📌 ✅ Answer:
⬥ Any rearrangement (shuffle) of digits keeps the same digit sum.
⬥ Therefore, any digit shuffle of a multiple of 9 is also a multiple of 9.

🔒 ❓ Q5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
📌 ✅ Answer:
⬥ A number divisible by 18 must be divisible by 2 and 9.

🟢 Step 1 ⬥ Divisibility by 2:
⬥ Last digit b must be even ⇒ b = 0, 2, 4, 6, 8

🟢 Step 2 ⬥ Divisibility by 9:
⬥ Digit sum = 4 + 8 + a + 2 + 3 + b = 17 + a + b
⬥ 17 + a + b must be a multiple of 9 ⇒ possible values = 18 or 27

⬥ For 18: a + b = 1
⬥ For 27: a + b = 10

🟡 Step 3 ⬥ Valid pairs (a, b):
⬥ (1, 0)
⬥ (2, 8), (4, 6), (6, 4), (8, 2)

➡️ All valid pairs: (1,0), (2,8), (4,6), (6,4), (8,2)

🔒 ❓ Q6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
📌 ✅ Answer:
⬥ 44 = 4 × 11 ⇒ number must be divisible by 4 and 11.
🟢 Step 1 ⬥ Divisibility by 4:
⬥ Last two digits = q8
⬥ q8 is divisible by 4 when q = 0, 2, 4, 6, 8
🟢 Step 2 ⬥ Divisibility by 11:
⬥ (3 + 7 + 8) − (p + q) = 18 − (p + q)
⬥ This must be a multiple of 11 ⇒ possible values = 0 or 11
⬥ 18 − (p + q) = 11 ⇒ p + q = 7
⬥ 18 − (p + q) = 0 ⇒ p + q = 18 (not possible)
🟡 Step 3 ⬥ Valid pairs:
⬥ (p, q) = (7,0), (5,2), (3,4), (1,6)

🔒 ❓ Q7. Find three consecutive numbers such that the first is a multiple of 2, the second a multiple of 3, and the third a multiple of 4. Are there more such numbers? How often do they occur?
📌 ✅ Answer:
⬥ Check numbers of the form n, n+1, n+2.

🟢 Step 1 ⬥ Smallest example:
⬥ 8 (multiple of 2), 9 (multiple of 3), 10 (not multiple of 4) ❌
⬥ 14, 15, 16 ✔️

⬥ 14 → divisible by 2
⬥ 15 → divisible by 3
⬥ 16 → divisible by 4

⬥ These occur every 12 numbers, since LCM(2,3,4) = 12.

➡️ Yes, more such numbers exist and they repeat every 12 numbers.

🔒 ❓ Q8. Write five multiples of 36 between 45,000 and 47,000. Share your approach.
📌 ✅ Answer:
⬥ 36 × 1250 = 45,000
⬥ Start from the next multiple.
⬥ Five multiples:
⬥ 45,036
⬥ 45,072
⬥ 45,108
⬥ 45,144
⬥ 45,180
⬥ Approach: find a nearby known multiple, then keep adding 36.

🔒 ❓ Q9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
📌 ✅ Answer:
◆ 5 consecutive even numbers differ by 2.
◆ Middle number = 5p.
◆ So the sequence is: (5p − 4), (5p − 2), 5p, (5p + 2), (5p + 4).
🔒 ❓ Q10. Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
📌 ✅ Answer:
◆ Example: 240045
◆ 240045 is divisible by 15 because it ends in 5 and (2+4+0+0+4+5)=15 is divisible by 3.
◆ Reversed number = 540042
◆ 540042 is divisible by 6 because it is even (ends in 2) and its digit sum is also 15 (divisible by 3).
🔒 ❓ Q11. 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.
📌 ✅ Answer:
◆ Let a multiple of 11 be 11k.
◆ Doubling it gives 2×(11k)=11×(2k), which is still a multiple of 11.
◆ So every multiple of 11 remains a multiple of 11 when doubled.
➡️ Deepak’s conjecture is false.

🔒 ❓ Q12. 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.
📌 ✅ Answer:
◆ (6a)(3b)=18ab=9(2ab).
➡️ Always True

🔒 ❓ (ii) The sum of three consecutive even numbers will be divisible by 6.
📌 ✅ Answer:
◆ Let the numbers be 2n, 2n+2, 2n+4.
◆ Sum = 6n+6 = 6(n+1).
➡️ Always True

🔒 ❓ (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
📌 ✅ Answer:
◆ Divisibility by 3 depends on digit sum; digit sum stays same after rearranging, so still divisible by 3.
◆ Divisibility by 2 depends on last digit; both numbers end with f, so evenness stays same.
➡️ Always True

🔒 ❓ (iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.
📌 ✅ Answer:
🟢 Step 1 ◆ Simplify the expression:
◆ 8(7b−3) − 4(11b+1)
◆ = (56b−24) − (44b+4)
◆ = 12b − 28
🟣 Step 2 ◆ Check divisibility by 12:
◆ 12b − 28 = 4(3b − 7) is always divisible by 4.
◆ But 3b − 7 is never divisible by 3 (since 3b is divisible by 3, and subtracting 7 leaves remainder 2).
➡️ Never True

🔒 ❓ Q13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
📌 ✅ Answer:
◆ Look at remainders when dividing by 3: 0, 1, 2.
◆ The sum is divisible by 3 exactly when the three remainders add to a multiple of 3.
◆ Possible remainder patterns:
◆ (0,0,0)
◆ (1,1,1)
◆ (2,2,2)
◆ (0,1,2) (in any order)
➡️ These are all the cases that make the sum divisible by 3.

🔒 ❓ Q14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these 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?
📌 ✅ Answer:
◆ Two consecutive integers: one must be even ⇒ product is always a multiple of 2.
◆ Three consecutive integers: among them, one is divisible by 3 and at least one is even ⇒ product is always a multiple of 6.
◆ Four consecutive integers:
◆ among them, at least one is divisible by 3, and there are two even numbers (one of which is a multiple of 4) ⇒ product always has factors 3 and 8 ⇒ multiple of 24.
◆ Five consecutive integers:
◆ contains a multiple of 5, a multiple of 3, and enough factors of 2 (from even numbers, including a multiple of 4) ⇒ product is always a multiple of 120.

🔒 ❓ Q15. Solve the cryptarithms —

🔒 ❓ (i) EF × E = GGG
📌 ✅ Answer:
◆ Solution: 37 × 3 = 111
◆ So E=3, F=7, G=1.

🔒 ❓ (ii) WOW × 5 = MEOW
📌 ✅ Answer:
◆ Solution: 575 × 5 = 2875
◆ So W=5, O=7, M=2, E=8.

🔒 ❓ Q16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
📌 ✅ Answer:
◆ Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4.
◆ So the sets are nested: Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4.
➡️ The correct diagram is the one with 32 inside 8 inside 4 (nested circles).

Class 8 : Maths – Lesson 5. Number Play

EXPLANATION AND ANALYSIS

TEXTBOOK QUESTIONS

OTHER IMPORTANT QUESTIONS

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