# 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. Find _6724.

Answer

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Hint: Find the sum of digits of the given number. Apply natural numbers from 0-9 in the blank space. A number is divisible by 3, if the sum of the digits are divisible by 3. Substitute different values in the sum and check divisibility.

Complete step-by-step answer:

We know that a number is divisible by 3, if the sum of the digits are divisible by 3.

We have to find the missing number in _6724.

The sum of the digits _6724 = _+6+7+2+4 = _+19.

Now we have to check different values for the blank digit and check if the number is divisible by 3.

Sum of digit = _+19

Let us put 0 to 9 digits in the blank space.

Let’s put variable ‘x’ in the blank space for easy calculation.

$\therefore $Sum of digit= x+19

Let’s put x=0, sum = 19+0=19

x=1, sum=19+1=20

x=2, sum=19+1=21

x=3, sum=19+1=22

x=4, sum=19+1=23

x=5, sum=19+1=24

x=6, sum=19+1=25

x=7, sum=19+1=26

x=8, sum=19+1=27

x=9, sum=19+1=28

From the above possibilities,

x=2, sum=21, is divisible by 3.

x=5, sum=24, is divisible by 3.

x=8, sum=27, is divisible by 3.

$\therefore $The smallest digit in blank space = 2.

The greatest digit in blank space =8.

Therefore the no. formed are 26724 and 86724

Note: Here we used to find a single digit to fit the blank space, that’s why the natural nos from 0-9 were considered.

Complete step-by-step answer:

We know that a number is divisible by 3, if the sum of the digits are divisible by 3.

We have to find the missing number in _6724.

The sum of the digits _6724 = _+6+7+2+4 = _+19.

Now we have to check different values for the blank digit and check if the number is divisible by 3.

Sum of digit = _+19

Let us put 0 to 9 digits in the blank space.

Let’s put variable ‘x’ in the blank space for easy calculation.

$\therefore $Sum of digit= x+19

Let’s put x=0, sum = 19+0=19

x=1, sum=19+1=20

x=2, sum=19+1=21

x=3, sum=19+1=22

x=4, sum=19+1=23

x=5, sum=19+1=24

x=6, sum=19+1=25

x=7, sum=19+1=26

x=8, sum=19+1=27

x=9, sum=19+1=28

From the above possibilities,

x=2, sum=21, is divisible by 3.

x=5, sum=24, is divisible by 3.

x=8, sum=27, is divisible by 3.

$\therefore $The smallest digit in blank space = 2.

The greatest digit in blank space =8.

Therefore the no. formed are 26724 and 86724

Note: Here we used to find a single digit to fit the blank space, that’s why the natural nos from 0-9 were considered.

Last updated date: 30th Sep 2023

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