NCERT solution class 8 chapter 16 Playing with numbers

Exercise 16.1   Exercise 16.2

Exercise 16.2

Question 1

If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Sol :

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 21y5 = 2 + 1 + y + 5 = 8 + y

Hence, 8 + y should be a multiple of 9.

This is possible when 8 + y is any one of these numbers 0, 9, 18, 27, and so on …

However, since y is a single digit number, this sum can be 9 only. Therefore, y should be 1 only.

 

 

Question 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?

Sol :

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z

Hence, 9 + z should be a multiple of 9.

This is possible when 9 + z is any one of these numbers 0, 9, 18, 27, and so on …

However, since z is a single digit number, this sum can be either 9 or 18. Therefore, z should be either 0 or 9.

 

 

Question 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)

Sol :

Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.

Sum of digits of 24x = 2 + 4 + x = 6 + x

Hence, 6 + x is a multiple of 3.

This is possible when 6 + x is any one of these numbers 0, 3, 6, 9, and so on …

Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus, x can have its value as any of the four different values 0, 3, 6, or 9.

 

 

Question 4

If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Sol :

Since 31z5 is a multiple of 3, the sum of its digits will be a multiple of 3.

That is, 3 + 1 + z + 5 = 9 + z is a multiple of 3.

This is possible when 9 + z is any one of 0, 3, 6, 9, 12, 15, 18, and so on …

Since z is a single digit number, the value of 9 + z can only be 9 or 12 or 15 or 18 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus, z can have its value as any one of the four different values 0, 3, 6, or 9.

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