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. Find _6724.

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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.

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.