Question

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

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Answer 1

Hey !

Here, Divisibility rule of 3 comes into picture.

Divisibility rule of 3:

It states that a number is divisible by 3 only when it's sum of the digits are divisible by 3.

Now,

a) __6724

Step-1 :

Find the sum of the digits of the number, assuming x as missing digit

=> x + 6+ 7 + 2 + 4

=> x + 19

Step-2:

Consider the nearest multiple to get the smallest digit

Nearest multiple of 3 after 19 is 21

For the sum of the digits to be equal to 21, find the value of missing digit

=> x+19 = 21

=> x = 21 - 19

=> x = 2

Step-3

Use trial and error Method, to get the greatest digit for missing digit

Consider sum of the digits again

=> x+19

Since, the greatest digit for a single digit is 9, we approach the numbers from 9 to 1

Digit --- Sum ----- Total ----- Divisible by 3

9 ------- 9+19 ------ 28 ------------- No

8 ------ 8+19 ------- 27 ------------- YES (9 times of 3)

so, the Greatest digit is 8

Answer :

(2)6724 - Filled Smallest digit

(8)6724 - Filled Greatest digit

Both are divisible by 3.

b)

4765__2

Step-1 :

Find the sum of the digits of the number, assuming x as missing digit

=> 4 + 7 + 6 + 5 + x + 2

=> x + 24

Step-2:

Here, 24 is itself divisible by 3, x can be zero as it is not the starting digit.

•°• Least digit = 0

Step-3

Use trial and error Method, to get the greatest digit for missing digit

Consider sum of the digits again

=> x+24

Since, the greatest digit for a single digit is 9, we approach the numbers from 9 to 1

Digit --- Sum ----- Total ----- Divisible by 3

9 ------- 9+24 ------ 33 ------------- YES (11 times of 3)

•°• Greatest digit = 9

Answer :

4765(0)2 - Filled Smallest digit

6765(9)2 - Filled Greatest digit

Both are divisible by 3

Meowwww xD

Answer 2

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Hey guys how r u ���

Here, Divisibility rule of 3 comes into picture.

Divisibility rule of 3:

It states that a number is divisible by 3 only when it's sum of the digits are divisible by 3.

Now,

a) __6724

Step-1 :

Find the sum of the digits of the number, assuming x as missing digit

=> x + 6+ 7 + 2 + 4

=> x + 19

Step-2:

Consider the nearest multiple to get the smallest digit

Nearest multiple of 3 after 19 is 21

For the sum of the digits to be equal to 21, find the value of missing digit

=> x+19 = 21

=> x = 21 - 19

=> x = 2

Step-3

Use trial and error Method, to get the greatest digit for missing digit

Consider sum of the digits again

=> x+19

Since, the greatest digit for a single digit is 9, we approach the numbers from 9 to 1

Digit --- Sum ----- Total ----- Divisible by 3

9 ------- 9+19 ------ 28 ------------- No

8 ------ 8+19 ------- 27 ------------- YES (9 times of 3)

so, the Greatest digit is 8

Answer :

(2)6724 - Filled Smallest digit

(8)6724 - Filled Greatest digit

Both are divisible by 3.

b)4765__2

Step-1 :

Find the sum of the digits of the number, assuming x as missing digit

=> 4 + 7 + 6 + 5 + x + 2

=> x + 24

Step-2:

Here, 24 is itself divisible by 3, x can be zero as it is not the starting digit.

•°• Least digit = 0

Step-3

Use trial and error Method, to get the greatest digit for missing digit

Consider sum of the digits again

=> x+24

Since, the greatest digit for a single digit is 9, we approach the numbers from 9 to 1

Digit --- Sum ----- Total ----- Divisible by 3

9 ------- 9+24 ------ 33 ------------- YES (11 times of 3)

•°• Greatest digit = 9

Answer :

4765(0)2 - Filled Smallest digit

6765(9)2 - Filled Greatest digit

Both are divisible by 3.

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