Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers Ex 16.2 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution:
21y5 is a multiple of 9 (given)
∴ Sum of digits of 21y5 = 2 + 1 + y + 5
= 8 + y
Therefore, (8 + y) should be 0, 9, 18, …, etc.
8 + y = 0 is not possible ∴ 8 + y = 9
∴ y = 9 – 8 = 1
Thus, the value of y is 1.
Verification:
21y5 = 2115 (∵ y = 1)
∴ Sum of digits of 2115 = 2 + 1 + 1 + 5 = 9 (9 ÷ 9 = 1, remainder = 0)
∴ 2115 is divisible by 9.
(Note: Here, verification is given to explain you.]
2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution:
31z5 is a multiple of 9. (given)
∴ Sum of digits of 31z5 = 3 + 1 + z + 5
= z + 9
Therefore, (z + 9) should be 0, 9, 18, …, etc.
Since, z is a digit, it should be either 0 or 9.
Hence, z = 0 or 9.
Verification:
31z5 = 3105 (∵ z = 0)
∴ Sum of digits of 3105
= 3 + 1 + 0 + 5 = 9
(9 ÷ 9 = 1, remainder = 0)
∴ 3105 is divisible by 9.
31z5 = 3195 (∵ z = 9)
∴ Sum of digits of 3195
= 3 + 1 + 9 + 5 = 18
(18 ÷ 9 = 2, remainder = 0)
∴ 3195 is divisible by 9.
3. If 24x is a multiple of 3, where x is a digit, what is the value of x? (Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 …………. But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution:
24x is a multiple of 3. (given)
∴ Sum of digits of 24x = 2 + 4 + x = 6 + x
Therefore, 6 + x should be 0, 3, 6, 9, 12, …, etc.
Since, 6 + x is a multiple of 3.
∴ 6 + x = 0, 6 + x = 3, 6 + x = 6, 6 + x = 9, 6 + x = 12, 6 + x = 15, 6 + x = 18, ……….
∴ x = -6, x = -3, x = 0, x = 3, x = 6, x = 9, x = 12, ……………..
Here, x = 0, 3, 6, 9 are possible.
Thus, the value of x can be 0 or 3 or 6 or 9.
4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
31z5 is a multiple of 3. (given)
∴ 31z5 is divisible by 3.
Sum of digits of 31z5 = 3 + 1 + z + 5
= 9 + z
∴ 9 + z is divisible by 3.
∴ Value of 9 + z should be 0, 3, 6, 9, 12, 15 or 18.
Since, z is a multiple of 3.
If 9 + z = 0,
∴ z = – 9 which is impossible.
9 + z = 3,
∴ z = – 6 which is impossible.
9 + z = 6,
∴ z = – 3 which is impossible.
9 + z = 9,
∴ z = 0 which is possible.
9 + z = 12,
∴ z = 3 which is possible.
9 + z = 15,
∴ z = 6 which is possible.
9 + z = 18,
∴ z = 9 which is possible.
9 + z = 21,
∴ z = 12 which is impossible.
Thus, the value of z can be 0 or 3 or 6 or 9.