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

AcademicMathematicsNCERTClass 6

To do :

We have to find the smallest digit and the greatest digit in the blank spaces so that the numbers formed are divisible by 3.

Solution :

Divisibility rule of 3:

Divisibility rule for 3 states that a number is completely divisible by 3 if the sum of its digits is divisible by 3.

(a) Let the digit in the blank space be $x$ where $0 \leq x \leq 9$.

x6724

Sum of the digits $= x + 6 + 7 + 2 + 4$

$= x + 19$

Now, $x+19$ is divisible by 3.

21 is the smallest multiple of 3 which comes after 19

This implies,

The smallest digit $x= 21 - 19$

$= 2$

The largest possible value of $x$ is,

$x+19$ is divisible by 3.

If $x=9, x+19=9+19=28$ which is not divisible by 3.

If $x=8, x+19=8+19=27$ which is divisible by 3.

Therefore,

The largest possible value of $x$ is 8.

Hence, the smallest digit is 2 and the greatest digit is 8.

(b) Let the digit in the blank space be $x$ where $0 \leq x \leq 9$.

4765x2

Sum of the digits $= 4 + 7 + 6 + 5 + x + 2$

$= x + 24$

Now, $x+24$ is divisible by 3.

In this case, 24 is the smallest multiple of 3.

This implies,

The smallest digit $x= 24 - 24$

$= 0$

The largest possible value of $x$ is,

$x+24$ is divisible by 3.

If $x=9, x+24=9+24=33$ which is divisible by 3.

Therefore,

The largest possible value of $x$ is 9.

Hence, the smallest digit is 0 and the greatest digit is 9.

raja
Updated on 10-Oct-2022 13:30:33

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