Question

Ex 3.3 Class 6 Maths 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​

Answer 1

Answer:

Solution:

We know that number is divisible by 3 if the sum of all the digits of the number is also divisible by 3.

(a) ___ 6724

Sum of the digits = 4 + 2 + 7 + 6 = 19

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

Hence, the required digits are 2 and 8.

(b) 4765 ____ 2.

Sum of digits = 2 + 5 + 6 + 7 + 4 = 24

The smallest digits to be place 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.

Hence, the required digits are 0 and 9.

Answer 2

$\star\:\:\:\bf\large\underline\blue{Question:-}$

Write the smallest and the greatest digit in the blank space of each of the following numbers so that the number formed in divisible by 3.

1. ___6724
2. 4765__2

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

For Question 1:-

We know that if the sum of the digits of the number is divisible by 3,then the number must be divisible by 3.

Here,

6+7+2+4=19.

Closest Number to 19 which is divisible by 3 is 21.

Difference=21-19=2.

So,

• Smallest number to be placed=2.

Now, if we put 9,then the sum will be=19+9=28.

But, 28 is not divisible by 3.

Also, 27 is divisible by 3 and 19+8=27.

• So, largest number that can be placed here=8.

For Question 2:-

Sum of digits = 4+7+6+5+2=24.

• Since, 24 is divisible by 3,therefore,smallest number that can be placed here = 0.

Now, if we write 9 at that place, the sum of digits will be =24+9=33 which is a multiple of 3.

Therefore, largest number that can be placed here = 9.