Class 8 Maths Ganita Prakash Chapter 5 Number Play NCERT Solutions

Class 8 Maths Ganita Prakash Chapter 5 Number Play NCERT Solutions

NCERT In-Text Question Page 122

Figure it Out

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

Solution:

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

Solution:

Question 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) The sum of two even numbers is a multiple of 3.

Claim: The sum of two even numbers is a multiple of 3.

Answer: ❌ Sometimes true

🔹 Algebraic justification

Let the two even numbers be:

• First even number = (2a)
• Second even number = (2b)

Their sum:

[2a + 2b = 2(a + b)]

This shows the sum is always even, but nothing guarantees it is divisible by 3.

For the sum to be a multiple of 3:

[2(a+b) must be divisible by 3]

Since 2 and 3 are coprime, this happens only when (a+b) is divisible by 3 — which is not always the case.

🔹 Example (true case)

Take:

• (6) and (12) (both even)

[6 + 12 = 18]

18 is a multiple of 3 ✔️

🔹 Non-example (false case)

Take:

• (4) and (8) (both even)

[4 + 8 = 12]

12 is a multiple of 3 (oops — need a better counterexample)

Try:

• (2) and (4)

[2 + 4 = 6]

Still multiple of 3 — try another.

Take:

• (2) and (8)

[2 + 8 = 10]

10 is not a multiple of 3 ❌

🔹 Conclusion

Because some pairs of even numbers give a multiple of 3 and others do not, the statement is:

[Sometimes true]

(ii) If a number is not divisible by 18, then it is also not divisible by 9.

Claim: If a number is not divisible by 18, then it is also not divisible by 9.

Answer: ❌ Sometimes true

🔹 Algebraic reasoning

Note that:

[18 = 2 x 9]

So every multiple of 18 is automatically a multiple of 9.
But the reverse is not always true.

Let a number divisible by 9 be:

[n = 9k]

For it to be divisible by 18, (k) must be even:

[n = 18m = 9(2m)]

If (k) is odd, the number is divisible by 9 but not by 18.

🔹 Counterexample (shows the statement is false)

Take:

[n = 9]

• 9 is divisible by 9 ✔️
• 9 is not divisible by 18 ❌

This contradicts the statement.

🔹 Example where the statement happens to be true

Take:

[n = 10]

• Not divisible by 18 ✔️
• Not divisible by 9 ✔️

So it works here — but not always.

🔹 Conclusion

Because there exist numbers (like 9, 27, 45, …) that are not divisible by 18 but are divisible by 9, the statement is:

[{Sometimes true}]

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

That statement isn’t always true. Let’s test it with some examples:

Step-by-step reasoning

• A number is divisible by 6 if it’s divisible by both 2 and 3.
• If a number is not divisible by 6, it could still be divisible by 2 or by 3, or by neither.
• The claim says: If two numbers are not divisible by 6, then their sum is not divisible by 6.
  Let’s check with counterexamples.

Counterexamples

1. Example 1:
   • Take 2 (not divisible by 6).
   • Take 4 (not divisible by 6).
   • Their sum = 6, which is divisible by 6.
     → Contradicts the statement.
2. Example 2:
   • Take 3 (not divisible by 6).
   • Take 9 (not divisible by 6).
   • Their sum = 12, which is divisible by 6.
     → Another contradiction.

Conclusion

The statement is false. Two numbers that are individually not divisible by 6 can still add up to a number that is divisible by 6.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

Let’s break this down carefully:

Step-by-step reasoning

• A multiple of 6 can be written as (6m), where (m) is an integer.
• A multiple of 9 can be written as (9n), where (n) is an integer.
• Their sum is: [ 6m + 9n ]
• Factor out 3: [ 6m + 9n = 3(2m + 3n) ]
• Since (2m + 3n) is an integer, the whole expression is a multiple of 3.

Conclusion

Yes, the statement is true:
The sum of a multiple of 6 and a multiple of 9 is always a multiple of 3.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Let’s test this statement carefully:

Step-by-step reasoning

• A multiple of 6 can be written as (6m), where (m) is an integer.
• A multiple of 3 can be written as (3n), where (n) is an integer.
• Their sum is: [ 6m + 3n ]
• Factor out 3: [ 6m + 3n = 3(2m + n) ]
• This shows the sum is always a multiple of 3.
• But for it to be a multiple of 9, we need: [ 6m + 3n = 9k \quad \text{for some integer } k ] which means (2m + n) must itself be a multiple of 3.

Counterexample

• Take (m = 1), (n = 1).
  Sum = (6(1) + 3(1) = 9). → divisible by 9.
• Take (m = 1), (n = 2).
  Sum = (6(1) + 3(2) = 12). → divisible by 3 but not divisible by 9.

Conclusion

The statement “The sum of a multiple of 6 and a multiple of 3 is a multiple of 9” is false.
It’s always a multiple of 3, but only sometimes a multiple of 9 (depending on whether (2m + n) is divisible by 3).

Question 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:

Question 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:

Question 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:

Answer: ✅ Always true

🔹 Algebraic proof

If a number leaves a remainder 2 when divided by 6, it can be written as:

[n = 6k + 2]

Let the three numbers be:

[6a + 2, 6b + 2, 6c + 2]

Add them:

[(6a+2) + (6b+2) + (6c+2)]

[= 6(a+b+c) + 6]

[= 6(a+b+c+1)]

This is clearly a multiple of 6.

🔹 Numerical example

Take numbers that give remainder 2 when divided by 6:

• 8, 14, 20

Check:

• (8 ÷ 6) remainder 2
• (14 ÷ 6) remainder 2
• (20 ÷ 6) remainder 2

Sum:

[8 + 14 + 20 = 42]

[42 ÷ 6 = 7]

✔️ Multiple of 6.

🔹 Why it always works (intuitive)

Each number contributes an extra +2.

Three such numbers contribute:

[2 + 2 + 2 = 6]

That extra 6 makes the total divisible by 6 every time.

✅ Conclusion

Tathagat’s claim is

[Always true]

because the sum of any three numbers of the form (6k+2) is always divisible by 6.

Question 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:

We are given:

661 = remainder 3 divided by7
4779 = remainder 5 divided by 7

(i) (4779 + 661)

🔹 Algebraic method

Write using remainders:

4779 = 7a + 5
661 = 7b + 3

Add:

(7a+5) + (7b+3)= 7(a+b)+8

Now reduce 8 divided by 7:

[8 = 7 + 1]

So remainder = 1

[{4779 + 661} = 1 divided by 7]

🔹 Visual (remainder) method

Think only about remainders:

• remainder of 4779 → 5
• remainder of 661 → 3

Add remainders:

[5 + 3 = 8]

Now divide 8 by 7:

• one group of 7
• remainder 1

✅ Remainder = 1

(ii) (4779 – 661)

🔹 Algebraic method

[(7a+5) – (7b+3) = 7(a-b) + 2]

So the remainder is 2.

[{4779 – 661} = 2 divided by 7]

🔹 Visual (remainder) method

Use remainders only:

• 4779 → remainder 5
• 661 → remainder 3

Subtract:

[5 – 3 = 2]

Already less than 7.

✅ Remainder = 2

✅ Final Answers

• (i) remainder 1
• (ii) remainder 2

Question 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:

Let the number be (n).

Given:

Number n When divided by 3 Remainder of 2
Number n When divided by 4 Remainder of 3
Number n divided When by 5 Remainder of 4

🔹 Key observation (the trick)

Notice:

[2=3-1, 3=4-1, 4=5-1]

So in each case,

[n (remainder) = -1 when divided by {3,4,5}]

This means:

[n+1 is divisible by 3,4,5]

🔹 Step 1: Use LCM

[{lcm}(3,4,5) = 60]

So:

[n+1 = 60k]

[n = 60k – 1]

🔹 Step 2: Smallest positive solution

Take smallest (k=1):

[n = 60 – 1 = 59]

🔹 Step 3: Verify

Check (59):

• (59 ÷ 3 = 19) remainder 2 ✔️
• (59 ÷ 4 = 14) remainder 3 ✔️
• (59 ÷ 5 = 11) remainder 4 ✔️

🔹 Visual intuition (why this works)

Each condition says the number is one less than a multiple:

• one less than a multiple of 3
• one less than a multiple of 4
• one less than a multiple of 5

So the number must be:

[{(common multiple of 3,4,5)} – 1]

The smallest common multiple is 60, so the smallest number is:

[60 – 1 = 59]

Any next solution would be:

[120 – 1 = 119, 180 – 1 = 179,—–]

which are larger.

✅ Final Answer

[{59}]

Why Smallest?
Because 60 is the smallest number divisible by 3, 4, and 5, and our number must be exactly one less than such a common multiple.

NCERT In-Text Question Page 126

Question 1: Find, without dividing, whether the following numbers are divisible by 9.

(i) 123
(ii) 405
(iii) 8888
(iv) 93547
(v) 358095

Solution:-

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

Solution:-

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

Solution:-

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

Solution:-

NCERT In-Text Question Page 126

Figure It Out

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

Solution:

Question 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:

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

Solution:

Question 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:

NCERT In-Text Question Page 132

Figure it Out

Question 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:-

Question 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:-

Question 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:-

Question 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:-

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

Solution:-

Note: We did not take total as 36, because a + b can not be more than 18 to make whole digits divisible by 9. If we take 36 total of a + b comes to 19 that is more than 18.

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

Solution:-

Question 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:-

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

Solution:-

Question 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:-

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

Solution:-

Question 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:-

Question 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) 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.

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

Solution:-

Question 14. 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?

Solution:-

Question 15. Solve the cryptarithms —

(i) EF × E = GGG
(ii) WOW × 5 = MEOW

Solution:

The solutions to the cryptarithms are:

(i) E = 3, F = 7, G = 1 (37 x 3 = 111)

(ii) W = 5, O = 7, M = 2, E = 8 (575 x 5 = 2875)

Case 1 Solution: Solve EF x E = GGG

Analyze the Produce: GGG represents a three-digit number with identical digits, which is a multiple of 111 (111, 222, 333, —- , 999).

Factorize 111: Since 111 = 3 x 37, any multiple GGG must be G x 3 x 37.

Match the Form EF x E:

The two-digit number EF must be a multiple of 37 becaue 37 is prime and cannot be formed by a single digit E.

Possibilities for EF: 37 or 74

Test Cases:

If EF = 37, then E = 3. Calculating 37 x 3 = 111. Here, G = 1. This satisfies the condition where E, F, and G are distinct.

If EF = 74, then E = 7. Calculating 74 x 7 = 518. This does not result in a GGG pattern.

(ii) Case 2 Solution: WOW × 5 = MEOW

Determine W: In the units place, W x 5 must end in W. This only happens if W = 0 or W = 5. Since W is the leading digit of WOW, W must be 5.

Set up the Equation: (500 + 10o + 5) x 5 = 1000M + 100E + 10o + 5.

Simplifying: 2525 + 50o = 1000M + 100E + 10o + 5.

Further simplifying: 2520 + 40o = 1000M + 100E.

Dividing by 10: 252 + 4o = 100M + 10E.

Solve for Digits:

The left side (252 + 4o) must be between 252 (if o = 0) and 288 (if o = 9).

The right side (100M + 10E) represents the two leading digits of the result. For the sum to be in the 250 — 290 range, M must be 2.

If M = 2, then 252 + 4o = 200 + 10E = 52 + 4o = 10E.

Testing o: if o = 7, then 52 + 28 = 80, so E = 8.

Verification: 575 x 5 = 2875, which matches MEOW where:

M = 2, E = 8, O = 7, W = 5.

Answer:

The final digit values are:

(i) E = 3, F = 7, G = 1
(ii) W = 5, O = 7, M = 2, E = 8

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

Solution:

(iv) Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64,…

Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64,….

Multiples of 32 are: 32, 64, 96, 128,…

The Venn diagram captures the relationship between the multiples of 4, 8, and 32: