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Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :
(a) 92 389 (b) 8 9484

Solution:

We will be using the concepts of divisibility by 11 to solve this.

(a) 92 __ 389

Let ‘x' be placed inside the blank.

Sum of its digits at odd places = 9 + 3 + 2

= 14

Sum of its digits at even places = 8 + x + 9

= 17 + x

Difference = 17 + x – 14

= 3 + a

The difference should be 0 or a multiple of 11, then the number is divisible by 11

If 3 + x = 0

x = - 3

But it cannot be a negative

Taking the closest multiple of 11 which is near to 3

It is 11 which is near to 3

Now, 3 + x = 11

x = 11 – 3

x = 8

Therefore the required digit is 8

(b) 8 __ 9484

Let ‘x’ be placed inside the blank.

Sum of its digits at odd places = 4 + 4 + x

= 8 + x

Sum of its digits at even places = 8 + 9 + 8

= 25

Difference = 25 – (8 + x)

= 17 – x

The difference should be 0 or a multiple of 11, then the number is divisible by 11

If 17 – a = 0

x = 17 ( not possible)

Now, let us take a multiple of 11.

considering as 11

17 – x = 11

x = 17 – 11

x = 6

Therefore, the required digit is 6.

You can also use the Online Divisibility Calculator to check this.

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.3 Question 6

Summary:

The digit in the blank space of each of the following numbers so that the number formed is divisible by 11 are 8 and 6 respectively.

☛ Related Questions:

Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 : (a) 92 __ 389 (b) 8 __ 9484