Question 1

Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and signs in between the numbers. How many different possibilities exist? Write all of them.
Evaluate each expression and write the result next to it. Do you notice anything interesting? (Page 112)

Accepted Answer

When four consecutive numbers are used with all possible combinations of ‘+’ and signs, the results will always be even. This is because, regardless of the sign placement, the sum will always involve adding and subtracting an equal number of consecutive numbers. Specifically, the sum of four consecutive numbers is always even, and the differences between them also result in even numbers when combined with plus and minus signs.
Example with 3, 4, 5, 6:
3 + 4 + 5 + 6 = 18
3 + 4 + 5 – 6 = 6
3 + 4 – 5 + 6 = 8
3 + 4 – 5 – 6 = -4
3 – 4 + 5 + 6 = 10
3 – 4 + 5 – 6 = -2
3 – 4 – 5 + 6 = 0
3 – 4 – 5 – 6 = -12
Observations:
All results are even numbers.
There are only eight possible combinations.
The results can be positive, negative, or zero.

Question 2

Now, take four other consecutive numbers. Place the ‘+’ and ‘-’ signs as you have done before. Find out the results of each expression. What do you observe? (Page 113)

Accepted Answer

Let us take numbers 5, 6, 7, 8:
Now,
5 + 6 + 7 + 8 = 26
5 + 6 + 7 – 8 = 10
5 + 6 – 7 + 8 = 12
5 + 6 – 7 – 8 = -4
5 – 6 + 7 + 8 = 14
5 – 6 + 7 – 8 = -2
5 – 6 – 7 + 8 = 0
5 – 6 – 7 – 8 = -16
Conclusion:
Regardless of the starting consecutive numbers, the same pattern of even results emerges when using all possible combinations of plus and minus signs. This is because the sum and difference of consecutive numbers will always result in an even number.
Intext Questions
Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers. (Page 115)
1. 2a + 2b
2. 3g + 5h
3. 4m + 2n
4. 2u – 4v
5. 13k – 5k
6. 6m – 3n
7. x
2
+ 2
8. b
2
+ 1
9. 4k × 3j
Solution:
1. Analyse the expression 2a + 2b
2a is even because it’s a multiple of 2.
2b is even because it’s a multiple of 2.
The sum of two even numbers is even, so 2a + 2b is always even.
2. Analyse the expression 3g + 5h
If g is odd and h is odd, then 3g is odd and 5h is odd.
The sum of two odd numbers is even,
e.g., 3 × 1 + 5 × 1 = 8.
If g is even and h is even, then 3g and 5h are even.
The sum of two even numbers is even,
e.g., 3 × 2 + 5 × 2 = 16.
If g is odd and h is even, then 3g is odd and 5h is even.
The sum of an odd and an even number is odd,
e.g., 3 × 1 + 5 × 2 = 13
Thus, 3g + 5h is not always even.
3. Analyse the expression 4m + 2n
4m is even because it’s a multiple of 2.
2n is even because it’s a multiple of 2.
The sum of two even numbers is even, so 4m + 2n is always even.
4. Analyse the expression 2u – 4v
2u is even because it’s a multiple of 2.
4v is even because it’s a multiple of 2.
The difference of two even numbers is even, so 2u – 4v is always even.
5. Analyse the expression 13k – 5k
This simplifies to 8k.
8k is even because it’s a multiple of 2.
Thus, 13k – 5k is always even.
6. Analyse the expression 6m – 3n
6m is always even.
3n can be odd (if n is odd) or even (if n is even).
If n is odd, 3n is odd, and the difference of an even and an odd number is odd,
e.g. 6 × 1 – 3 × 1 = 3.
Thus, 6m – 3n is not always even.
7. Analyse the expression x
2
+ 2
x
2
is odd or even, both.
If x
2
is even, x
2
+ 2 becomes even.
If x
2
is odd, x
2
+ 2 becomes odd.
Thus, x
2
+ 2 is not always even.
8. If b is even, b2 is even, and b
2
+ 1 odd,
e.g., 2
2
+ 1 = 5.
If b is odd, b
2
is odd, and b
2
+ 1 is even,
e.g, 3
2
+ 1 = 10.
Thus, b
2
+ 1 is not always even.
9. Analyse the expression 4k × 3j
This simplifies to 12kj.
12kj is even because it’s a multiple of 2.
Thus, 4k + 3j is always even.
Figure It Out (Pages 122-123)

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

Accepted Answer

(i) Sometimes true, the sum of two even numbers is a multiple of 3.
Examples:
2 + 4 = 6, 8 + 10 = 18, 14 + 16 = 30 are multiples of 3.
2 + 6 = 8, 4 + 10 = 14 are not multiples of 3.
(ii) Sometimes true, if a number is not divisible by 18, then it is also not divisible by 9.
Examples:
27 is not divisible by 18, but 27 is divisible by 9. True
40 is not divisible by 18, also it is not divisible by 9. False
(iii) Never true, if two numbers are not divisible by 6, then their sum is not divisible by 6.
Examples:
8 and 10 are not divisible by 6.
The sum of two numbers = 8 + 10 = 18, is divisible by 6.
10 and 13 are not divisible by 6.
The sum of 10 and 13 = 10 + 13 = 23, which is not divisible by 6.
(iv) Always true
Multiple of 6; 6m
Multiple of 9; 9n
6m + 9n = 3(2m + 3n)
Hence multiple of 3.
(v) Sometimes true, the sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Multiples of 6 are: 6, 12, 18, 24, 30,…
6 + 12 = 18 is a multiple of 9.
12 + 18 = 30 is not a multiple of 9.
18 + 24 = 42 is not a multiple of 9.
Sometimes true.
Multiples of 3 are: 3, 6, 9, 12, 15, 18,…
3 + 6 = 9 is a multiple of 9.
6 + 9 = 15 is not a multiple of 9.
Sometimes true.

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.

Accepted Answer

Here, Remainder = 2, Dividend = 3
∴ Number = (Quotient × Dividend) + Remainder = (K × 3) + 2
where, K = 1, 2, 3,…..
Numbers = 1 × 3 + 2 = 3 + 2 = 5
Numbers = 2 × 3 + 2 = 6 + 2 = 8
Numbers = 3 × 3 + 2 = 9 + 2 = 11
Thus, 5, 8, and 11 are numbers that leave a remainder of 2 when divided by 3.
Algebraic expression = 3K + 2
Here, Remainder = 2, dividend = 4
Number = 4K + 2, where K = 1, 2, 3, 4, …
Numbers = 4 × 1 + 2 = 4 + 2 = 6
Numbers = 4 × 2 + 2 = 8 + 2 = 10
Numbers = 4 × 3 + 2 = 12 + 2 = 14
Algebraic expression = 4K + 2
Thus, 6, 10, and 14 are numbers that leave a remainder of 2 when divided by 4.

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?”

Accepted Answer

The LCM of 3, 5, and 7 = 3 × 5 × 7 = 105 [∵ 3, 5, and 7 are prime numbers]
No. of pebbles = 105 + 1 = 106

Source: This question is from Number Play, Mathematics &mdash; Class 8, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Number Play: Hold, Pebbles, Too, Group, Stays, Try. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 8 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
Related Questions from Number PlayQ1: Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and signs in between the...
Q10: Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is...
Q11: Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But...
Q12: Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’....
Q13: Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Practice more questions from Number Play &mdash; Mathematics, Class 8 CBSE

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?

Accepted Answer

The expression has been written by Tathagat = 6k + 2
where, k = 1, 2, 3, 4, 5, 6,…
6 × 1 + 2 = 8
6 × 2 + 2 = 14
6 × 3 + 2 = 20
6 × 4 + 2 = 26
The sum of three numbers
8 + 14 + 20 = 42, it is a multiple of 6.
14 + 20 + 26 = 60, it is a multiple of 6.
Yes, Tathagat’s claim is true.

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 77? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 – 661

Accepted Answer

Given, 661 = K × 7 + 3, where K = 1, 2, 3, 4, …
and, 4779 = K × 7 + 5
Algebraic Method:
(i) 4779 + 661
4779 = (682 × 7 ) + 5
Remainder = 5
661 = (94 × 7) + 3
Remainder = 3
∴ $\frac{4779+661}{7}=\frac{5+3}{7}=\frac{8}{7}$ = 1 = Remainder
(ii) 4779 – 661
∴ $\frac{4779-661}{7}=\frac{5-1}{7}=\frac{8}{7}$ = 4 = Remainder
Visualization Method:
(i) 4779 + 661 = (682 × 7) + 5 + (94 × 7) + 3
= 7 × (682 + 94) + 5 + 3
= 7 × 776 + 8
= Divisible by 7 + $\frac {8}{7}$
= 1, Remainder
(ii) 4779 – 661 = (682 × 7) + 5 – (94 × 7) – 3
= 7 × (682 – 94) + 5 – 3
= 7 × 588 + 2
= Divisible by 7 + 2
= 2, Remainder

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?

Accepted Answer

The expression of a number that leaves a remainder of 2 when divided by 3.
Number = 3K + 2
3 × 1 + 2 = 5
3 × 2 + 2 = 8
3 × 3 + 2 = 11
3 × 4 + 2 = 14
3 × 5 + 2 = 17
3 × 6 + 2 = 20
The expression of a number that leaves a remainder of 3 when divided by 4.
Number = 4K + 3
4 × 1 + 3 = 7
4 × 2 + 3 = 11
4 × 3 + 3 = 15
4 × 4 + 3 = 19
The expression of a number that leaves a remainder of 4 when divided by 5.
Number = 5K + 4
5 × 1 + 4 = 9
5 × 2 + 4 = 14
5 × 3 + 4 = 19
5 × 4 + 4 = 24
Smallest number = LCM of (3, 4, 5) – 1
= 60 – 1
= 59
59 is the smallest 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.
5.2 Checking Divisibility Quickly
Figure It Out (Page 126)

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.

Accepted Answer

Given the middle number in the sequence of 5 consecutive even numbers 5p.
The other four numbers in the sequence in terms of p are 5p – 4, 5p – 2, 5p + 2, 5p + 4
Hence, the other four numbers in sequence are p, 3p, 7p, and 9p.

Question 10

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

Accepted Answer

We know that if the number is divisible by 3 and 5, then it is also divisible by 15.
Consider the number 643215.
Sum of the digits = 6 + 4 + 3 + 2 + 1 + 5 = 21, which is divisible by 3.
Thus, 643215 is divisible by 3.
One’s place = 5, it is also divisible by 5.
Hence, 643215 is divisible by 15.
One’s place is not 0, because the digits are reversed, it becomes a 5-digit number.
Lakhs place is always taken as an even number.
Reversed the digits:
512346
One’s place = 6, 512346 is divisible by 2.
Sum of the digits = 5 + 1 + 2 + 3 + 4 + 6 = 21.
It is also divisible by 3.
Hence, 512346 is divisible by 6.

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.

Accepted Answer

The multiples of 11 are: 11, 22, 33, 44, 55,…
When doubled, 22, 44, 66, 88, 110,……
i.e. (11) × 2, 11 × 4, 11 × 6, 11 × 8, 11 × 10,…. are also multiples of 11.
False, if multiples of 11 are doubled, then the multiples of 11 are these numbers.

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.

Accepted Answer

(i) Always True,
The multiple of 6 can be written as 6a, where a is an integer.
The multiple of 3 can be written as 3b, where b is an integer.
∴ Product = (6a) × (3b) = 18(ab) is a multiple of 9.
(ii) Always True,
The sum of three consecutive even numbers will be divisible by 6.
For example 2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 + 10 + 12 = 30,…
These numbers are divisible by 6.
(iii) Always True, because one’s place does not change.
(iv) Sometimes true,
Conclusion:
8(7 × 1 – 3) – 4(11 × 1 + 1) = -16, not divisible by 12.
8(7 × 10 – 3) – 4(4 × 10 + 1) = 536 – 164 = 372, divisible by 12.

Question 13

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

Accepted Answer

Let the three numbers be n
1
, n
2
, and n
3
.
Let their remainders when divided by 3 be r
1
, r
2
, and r
3
.
The sum n
1
+ n
2
+ n
3
is divisible by 3 if and only if r
1
+ r
2
+ r
3
is divisible by 3.
Case 1: All remainders are 0.
r
1
= 0, r
2
= 0, r
3
= 0
Sum of remainders = 0 + 0 + 0 = 0, which is divisible by 3.
Case 2: All remainders are 1.
r
1
= 1, r
2
= 1, r
3
= 1
Sum of remainders = 1 + 1 + 1 = 3, which is divisible by 3.
Case 3: All remainders are 2.
r
1
= 2, r
2
= 2, r
3
= 2
Sum of remainders = 2 + 2 + 2 = 6, which is divisible by 3.
Case 4: One remainder is 0, one is 1, and one is 2.
r
1
= 0, r
2
= 1, r
3
= 2 (in any order).
Sum of remainders = 0 + 1 + 2 = 3, which is divisible by 3.
The sum of three numbers is divisible by 3 if and only if all three numbers have the same remainder when divided by 3, or if they all have different remainders when divided by 3.

Question 14

Is the product of two consecutive integers always a 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?

Accepted Answer

Yes, the product of two consecutive integers is always a multiple of 2.
1 × 2 = 2, 2 × 3 = 6, 5 × 6 = 30, 10 × 11 = 110, and so on.
Since we know that multiplying by an odd number and an even number is always an even number.
No, it is not always a multiple of 6.
1 × 2 = 2, 4 × 5 = 20, 7 × 8 = 56
Since it is not divisible by 6.
The product of 4 consecutive integers
2 × 3 × 4 × 5 = 120,
4 × 5 × 6 × 7 = 840,
5 × 6 × 7 × 8 = 1680
We can say that the product of 4 consecutive integers, divisible by 12.
The product of five consecutive integers is:
1 × 2 × 3 × 4 × 5 = 120,
2 × 3 × 4 × 5 × 6 = 720,
3 × 4 × 5 × 6 × 7 = 2520
Hence, we can say that the product of five consecutive integers is always divisible by 24.

Question 15

Solve the cryptarithms
(i) EF × E = GGG
(ii) WOW × 5 = MEOW

Accepted Answer

(i) This means a 2-digit number multiplied by 5 gives a 3-digit number.
2-digit number = 20, 21,…., 99
37 × 3 = 111, all conditions are satisfied.
(ii) This means a 3-digit number multiplied by 5 gives 4-digit numbers.
Pick 3-digit number = 200, 201,…., 999
525 × 5 = 2625

Question 16

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

Accepted Answer

(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: