Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3
(a) ____ 6724   (b) 4765 ____ 2

Solution:

We will be using the concepts of divisibility by 3 to solve this.
A number is divisible by 3 if the sum of all digits in the number is also divisible by 3.

(a) ___ 6724
Sum of the digits = 4 + 2 + 7 + 6 = 19
Thus, The smallest digit to be placed is blank space = 2.
Then the sum = 19 + 2 = 21, which is divisible by 3.
The greatest digit to be placed in blank space = 8.
Then, the sum = 19 + 8 = 27, which is divisible by 3
Therefore, the required digits are 2 and 8.

(b) 4765 ____ 2.
Sum of digits = 2 + 5 + 6 + 7 + 4 = 24
Thus, The smallest digits to be placed in blank space = 0.
Then, sum = 24 + 0 = 24, which is divisible by 3.
The greatest digit to be placed in blank space = 9.
Then, the sum = 24 + 9 = 33, which is divisible by 3.
Therefore, the required digits are 0 and 9.

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

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.3 Question 5

Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3. (a) ____ 6724 (b) 4765 ____ 2

Summary:

The smallest digit and the greatest digit so that the number formed is divisible by 3 are (a) 2 and 8 (b) 0 and 9

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