Class 8 Math Ch 16 Playing With Numbers Ex 16.2

In Class 8 Math Ch 16 Playing with Numbers Ex 16.2, students continue to explore important concepts related to numbers, such as factors, multiples, and divisibility rules. This exercise helps strengthen understanding of how numbers interact through these properties and how to solve problems involving them.

By practicing Ex 16.2, students develop critical thinking skills and improve their ability to work with number patterns, which are essential for more advanced mathematics. This chapter lays the groundwork for topics like algebra and helps students build confidence in handling numerical problems.

Class 8 Math Ch 16 Playing with Numbers Ex 16.2 – Textbook

Ex 16.2 Class 8 Maths Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Solution:
A number is divisible by 9 if the sum of its digits is also divisible by 9.
Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y
(8 + y) ÷ 9 = 1
⇒ 8 + y = 9
⇒ y = 9 – 8 = 1
Hence, the required value of y = 1.

Ex 16.2 Class 8 Maths Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

Solution:
A number is a multiple of 9 when the sum of its digits is also divisible by 9.
Sum of the digits of 31z5 = 3 + 1 + z + 5
3 + 1 + 2 + 5 = 9k where k is an integer.
For k = 1,
3 + 1 + z + 5 = 9
⇒ z = 9 – 9 = 0
For k = 2,
3 + 1 + z + 5 = 18
⇒ z = 18 – 9 = 9
k = 3 is not possible because 3 + 1 + z + 5 = 27
⇒ z = 27 – 9 = 18 which is not a digit.
Hence the required value of z is 0 or 9

Ex 16.2 Class 8 Maths Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?

Solution:
Since 24x is a multiple of 3, the sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers; 0, 3, 6, 12, 15, 18, ……..
6 + x = 3k where k is any integer.
For k = 0,
6 + x = 3 × 0
⇒ 6 + x = 0
x = -6. Not possible
For k = 1,
6 + x = 3 × 1
⇒ 6 + x = 3
⇒ x = 3 – 6 = -3. Not possible
For k = 2,
6 + x = 3 × 2
⇒ 6 + x = 6
⇒ x = 6 – 6 = 0
2 + 4 + 0 = 6 multiple of 3
For k = 3,
6 + x = 3 × 3
⇒ x = 9 – 6 = 3
2 + 4 + 3 = 9 multiple of 3

For k = 4,
6 + x = 3 × 4
⇒ 6 + x = 12
⇒ x = 12 – 6 = 6
2 + 4 + 6 = 12 which is multiple of 3

For k = 5,
6 + x = 3 × 5
⇒ x = 15 – 6 = 9
2 + 4 + 9 = 15 which is multiple of 3
For k = 6,
6 + x = 3 × 6
⇒ x = 18 – 6 = 12 not possible as x is digit
Hence the required values of x are 0, 3, 6 or 9.

Ex 16.2 Class 8 Maths Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the value of z?

Solution:
A number is a multiple of 3 if the sum of its digits is divisible by 3.
3 + 1 + z + 5 = 3k where k is an integer
⇒ 9 + z = 3k
⇒ z = 3k – 9
Here, k = 0, 1, 2 is not possible as z is a digit of the number.

For k = 3,
z = 3 × 3 – 9 = 9 – 9 = 0
9 + 0 = 9 multiple of 3

For k = 4,
z = 3 × 4 – 9 = 12 – 9 = 3
9 + 3 = 12 multiple of 3

For k = 5,
z = 3 × 5 – 9 = 15 – 9 = 6
9 + 6 = 15 multiple of 3
For k = 6,
z = 3 × 6 – 9 = 18 – 9 = 9
9 + 9 = 18 multiple of 3

For k = 7,
z = 3 × 7 – 9 = 21 – 9 = 12 not possible as z is a digit
Hence, the required values of 2 are 0, 3, 6 and 9.

Class 8 math Ex 16.3

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Class 8 Math Ch 16 Playing with Numbers Ex 16.2 is an important exercise that deepens students’ understanding of number properties, including factors, multiples, and divisibility rules.

By practicing this exercise, students enhance their logical reasoning and problem-solving abilities, which are essential for tackling more complex math problems in the future.

Moreover, Class 8 Math Ch 16 Playing with Numbers Ex 16.2 lays a strong foundation for algebra and other higher-level mathematics topics by strengthening students’ grasp of basic number concepts.

Consistent practice of this exercise helps students build confidence and accuracy when working with numbers. Overall, Class 8 Math Ch 16 Playing with Numbers Ex 16.2 is a valuable resource for Class 8 learners aiming to master fundamental math skills.

For the official Class 8 Mathematics Solutions, you can visit:

NCERT Textbooks (for Class 8):

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